 So today's session I would be starting with harmonic progression, harmonic progression. So this is the third progression that is of importance to us. We had already covered arithmetic progression, arithmetic mean, harmonic, sorry, geometric progression, geometric mean, only harmonic progression is left to be, you know, addressed. Don't worry, it's a very simple type of progression. So what's a harmonic progression? Harmonic progressions are basically a sequence where the reciprocal of the terms of that sequence becomes a arithmetic progression. So if you see, this is a sequence that I've listed in front of you. If you start reciprocating all the terms, okay? So let's say I call this as a T1, okay? And I call this as a T2, I call this as a T3 and da-da-da-da-da. Let's say this is your Tn. So if you see the reciprocal of these terms, okay, 1 by T1, 1 by T2, 1 by T3, etc., they will start behaving as, they will start behaving as an arithmetic progression. So if this is in HP, then the reciprocal of these terms will be in AP, okay? That is to say that the difference of the reciprocal of these terms would be a constant. In this case, I've taken it as a D. So this is, this will be a constant, okay? So in plain simple and simple language without any sugarcoating on it, what is a harmonic progression? Harmonic progression is nothing but terms whose reciprocals, or if you reciprocate all the terms of a harmonic progression, you get an arithmetic progression. Is that fine? Okay. So many times when harmonic progression questions are asked, what do we do? We just reciprocate all the terms and we basically start using our arithmetic progression ideas or concerns on that, okay? Now the good or the bad part of harmonic progression is there is no direct formula to find the sum of n terms of a harmonic progression, right? So there is no direct formula to find the sum of n terms of a harmonic progression, okay? So we will not be going into that part. So what kind of questions? What kind of concepts will be tested on harmonic progression? So let us begin with some questions. You will automatically get an idea. What are the type of questions or what are the type of concepts that you will be asked? Okay. So let me begin with a simple question, okay? So let me start with this question. Find the first term of a HP whose second term is five by four and third term is half, okay? So find the first term of a HP whose second term is five by four and third term is a half. Please give me a response on the chat box. If you are done, please feel free to give me a response on the chat box. Very good, Nvesh. Some of you were absent during the examination time. Okay, by the way, how was your paper? How was your school semester exam? Have you started getting your marks? If yes, kindly share your scores with us. Don't worry, I'll not take your name. Okay, Bashnav. So NPS Kormangala, have you started getting your papers? Okay, once you go to the school next week, you will get your papers. Okay, what about NPS HSR and others? For that matter, any school, you come from any school. Have you got your marks? If yes, you can please share your score with us. So should we discuss this? Very good, Tejaswini. Siddharth, very good. Anybody else who wants to answer this? Okay, so let's say I have three terms of an HP like this. One by A, one by A plus D and one by A plus 2B. So they're asking us, what is this one by A? Given that the second term is five by four and third term is a half. That is to say A plus D is four by five and A plus 2D is a two. Okay, so let me call this as a one. Let me call this as a two. Just subtract two minus one. So that'll give you D is equal to two minus four by five. That's nothing but six by five. Okay, so once you've got D, you can use any of the one and two equations to get your A. So this means A plus six by five is equal to four by five. That means A is equal to negative two by five. But remember, A is not our first term. Our first term is one by A. So if you reciprocate that, you will end up getting minus five by two as your first term. So well done. Those who got minus five by two, that's absolutely right. Is it fine? Any questions, any concerns? Pretty straightforward, simple question. Can we move on to the next one already? Okay, let's go to the next one. So next one we have, okay, this question, which is actually a important property as well. Question says, if A, B, C are in HP, then show that A minus B by B minus C is equal to A by C. Okay, in fact, I will extend this in a property to AP and GP as well. So please note down the following. If A, B, C are in AP, okay? And we'll prove these as well, don't worry. Then please remember A minus B by B minus C will be A by A, okay? If A, B, C are in GP, are in GP, then please remember A minus B by B minus C is equal to A by B. And lastly, the question itself, if A, B, C are in HP, then please note down A minus B by B minus C is equal to A by C, which is the question itself, okay? So I just extended this property to AP and GP as well, okay? So instead of doing the question only, we'll be trying to prove all these three over here. They're very simple to prove. Let's start with the very first one that I wrote here. If A, B, C are in AP, then prove that A minus B by B minus C will be equal to A by A. Okay, now that is very simple. You'll say that's obvious because A minus B by B minus C will become equal to one, which implies A minus B is equal to B minus C, which means A plus C is equal to two B, which is actually the characteristic of an arithmetic progression. So in an arithmetic progression, we have already learned in our last class that some of the extreme two terms is twice of the middle. Okay, so if A, B, C are in AP, A plus C is equal to two B, and therefore, yes, this property is very much valid. Okay, so let me write it down, which is true for A, B, C in AP. Is there fine any questions with respect to the first one? Second one, similarly, let's take that up. Second one, we have to prove that A minus B by B minus C. So let me just cross-multiply. So let me multiply this with a B is equal to A times B minus C. So this means A, B minus B square is equal to A, B minus A, C. A, B, A, B goes for a toss, and that's as good as saying B square is equal to A, C, which is again true for A, B, C being in GP. So if A, B, C are in GP, we already know from our last-class discussion on geometric progression that B square will be equal to A, C. Okay, now coming to the last part of this particular proof, which is actually your question. So let us understand here one thing before I solve this. If A, B, C are in HP, then that means one by A, one by B, and one by C are in AP. That means to say twice of the middle is sum of the extremes. Correct? So can I get this result from this? I mean, do these two convey the same meaning? Are they the same? Let's try to figure it out. Okay, so yeah. Okay, so Siddharth is saying, yes, they are same. Okay, let's check it out, Siddharth. So here what I'm going to do, again, I'm going to do a cross-multiplication. So C times A minus B is A times B minus C. So when you open this bracket, you get AC minus BC is equal to AB minus AC. One very important thing I would like to say here, even though it may sound very trivial, whenever you are expanding terms, which contains alphabets, always write them in alphabetical way. For example, AC, don't write CA. Okay, BC, don't write CB. Why I do suggest that even though it's very trivial, is because it helps you to recognize the terms which you can cancel off. Okay, or you can basically associate. Okay, if you start writing AC somewhere and CA somewhere, your mind itself will get confused whether they are same terms or not. Okay, so it's always a solution. Same with XYZ also. Okay, so if you have alphabets, always write them in alphabetical order. It is for your own, you can say convenience. So this gives you two AC as AB plus BC. Now do one thing, divide throughout with ABC. Divide throughout with ABC. Now please note that if three terms are in harmonic progression, none of them can be zero. Okay, none of them can be zero. Okay, so when you do this, you end up getting two by B is equal to one by C plus one by A. And that's what we had written over here. Please check. Okay, so this is what was written here at the left side of your screen, right? So if three numbers are satisfying this characteristic, that means, yes, truly they are in HP, is it fine? So just keep this in mind. Why have I given these properties to you? Because sometimes a direct question may be framed in the regional entrance exams. Not J and J advance because it is too simple for J or J advance to ask such questions. But yes, if you know them, you save some time in these kinds of questions. Is it fine, any questions, any concerns that you want to get addressed over here before I move on to the next slide? Sir? Yes, yes, yes, please tell me. Sir, what is ABC written down there? Oh, Akaansh is Abhijay. So what is ABC written down below two ACs equal to AB plus BC? Where, where, where? ABC, I divided by ABC. I divided by ABC. Right, Abhijay? Yes, sir. Okay. You have logged in from a different name, so I was confused, who is that? Is this fine? Any questions? All right. So let's move on to the next problem. Let's move on to the next problem. I will not be taking a lot of questions, don't worry. Okay, let's, let's take a question where there is a use of AP, GP, HP in, in combination. So there are many cases where you realize that the question setter has given a mixed back question where AP, GP, HP, et cetera are involved. So this is one of those questions. It says, if ABC are in AP and A square, B square, C square, B in HP, then prove that either minus A by two comma C are in GP or ABC themselves are equal. Please attempt this question. And in case you have proved it, do write it down on the chat box. So if ABC are in AP, the first thing information that we have at hand is that two B is equal to A plus C, correct? And if one by, sorry, if A square, B square, C square are in HP, then we can also say that two by B square is equal to one by A square plus one by C square, okay? By the way, a nice way to write this is also B square is equal to two A square C square by A square plus C square. So this is something which I would like to, you to note down very carefully. If three numbers are in HP, if three numbers are in HP, then please remember the middle number is twice XZ by X plus Z, okay? It's the same thing as saying two by Y is equal to one by X plus one by Z, okay? So this is something which you can keep in your mind. It is very handy while solving questions. So in the same way, I can say B square is equal to two A square C square by A square plus C square, okay? Now I will stop over here. I'll give you some time to think over it. So how do we use one and two to get to the fact that either these three are in GP or ABC are equal? I hope all of you would have seen our JEE Advanced Results which came out on October 15th. And you can observe there that the very same people who did well in JEE main, of course, they also did well in KVPY. They are the ones who did well in their JEE Advanced Exam. And I'm very sure. I mean, they also did well in their Boards Exam as well which whose results came out much earlier. So basically what I want to say is that if you are clear with your concepts, if you love solving problems, if you like challenges, if you like learning, so as to say, it doesn't matter which exam you are writing, right? There's no special preparation for Boards or JEE. You just have to fall in love with learning. If you do that, you will ace any exam that you aspire for, okay? So same set of people have aced all the exams, right? So I think four of them are under, I mean, very close to a thousand ranks. I think one of them is from Indanagar batch. Ruchir Singh is from HSR. You must be knowing Ruchir Singh as well. Two of them are from NPS Raja Ji Nagar. I think three of them are from NPS Raja Ji Nagar, okay? And few of them are from non-NPS schools as well. If possible, and you want to speak to these toppers, I can share their number with you, of course, after taking permission from them. Please speak to them because they will give you a firsthand experience of how they have managed these two years of preparation. In fact, their preparation was even lengthier. It was two years, almost 10 months because the exam happened in October. So it was more painful for them too, because the preparation was prolonged. So do speak to them. It'll give you a great, they'll give you some insights about how to go about with balancing school versus preparation. Because most of you need it, right? Because I can see temporary absence during the exams. They were never absent during school exams. Let me tell you very frankly, I mean, you will say, sir, you are saying that because you want us to be present. No, talk to them. They will tell you. Okay, let's try to solve this question. Anybody who could get success here? Anybody who could solve it? Okay, so let's do this. From the second expression, let's write this. So from two, I am writing it like this. Okay. Now, what I plan to do is I plan to write this a square plus c square as a plus c the whole square. Okay. Minus two ac, okay? A plus c is already two b. So let me write, let me write this as a two b square. Two b square will be four b square, isn't it? If possible, let's drop the factor of, let's drop the factor of two from both the sides. So let me drop a factor of two from both the sides. So that'll give you something like this, okay? Let's expand it. So that gives us two b to the power four minus b square ac. And let's bring this term. Let me write it like this, a b square c. And let me bring this term to the left-hand side, okay? Is this factorizable? Is this factorizable? Let's try to figure it out if it is factorizable. Let's write this term as negative two a b square c plus a b square c. Oh yes, I think it's going to be factorizable. So take two b square common, it just gives you b square minus ac, correct? And from here, if you take ac common, again, it gives you b square minus ac. So overall it is factorizable as this, okay? Now this implies two things for us. This implies two things for us. Either two b square is negative ac, which means b square is negative a by two c, okay? Which means minus a by two b and c are in GP. And that basically proves the first part of the question where we had asked to prove this, okay? So minus a by two b c are in GP. And the second part says b square should be equal to ac, which anyways means a b c are in GP, okay? Now, here's an important theory which I would like to share with you. If three numbers, please note this down, very important part. If three numbers a b c are in any of the two progressions, in any of the two progressions, okay? Then they will also be, they will also be in the third progression as well, okay? And not only that, they would be equal as well. And they would be equal as well, okay? And they would be equal as well, okay? So what does this mean? Let me explain this once again. So it says that if three numbers are in any of the two progression, then it will also be in the third progression. Let's say if a b c are in AP and a b c are in GP, okay? Then a b c will also be in HP for sure, right? So any of the two progression they are in, then the third one, they will also be automatically. So let's say they're in AP and HP both, then they will be in GP also. If they're in GP and HP both, then they will be in AP also, okay? And not only that, they will all be equal, hence they will be in all the three, okay? So three equal numbers are in AP also, GP also, HP also, isn't it? So this is the fact which I'm going to use over here. So if a b c are in GP and it is already given that a b c are in AP, that means a b c should be equal. And of course they will be in HP as well. So a b c will be in HP as well. And not only that, they will all be equal also. And that's what the question center wants us to prove, okay? So both the proofs you have done, that is minus a by two b and c are in GP and a b c are all equal as well. Is it fine? Any questions? Please note down if you have any questions, any queries, any concerns, do let me know, okay? All right. So we are going to not talk about harmonic mean. So I will talk about two things over here. One is harmonic mean off. Harmonic mean off, A1, A2, A3, et cetera, which are non-zero numbers. So these are non-zero quantities. Please note that as I already told you, no term in a harmonic progression should be zero, okay? So if A1, A2, A3, An are non-zero quantities, what is the harmonic mean off? Now the word here is off, okay? So harmonic mean off non-zero quantities is defined as, please note this down, total number of terms divided by some of their reciprocals, okay? Da-da-da-da, till one by An, okay? So total number of terms divided by some of their reciprocals, okay? What is important to us is the second concept which is associated with harmonic mean, which is insertion of harmonic mean between insertion of harmonic mean or means between two non-zero quantities, okay? Between two non-zero quantities, two non-zero quantities, okay? Non-zero numbers, you can say quantities is as good as saying non-zero numbers, okay? This concept is more important. So let us say I have two non-zero numbers A and B and I want to insert one harmonic mean between these numbers, okay? So what is H1 here in terms of A and B? Please write down your response on the chat box. Okay, write down H1 in terms of A and B. So first of all, what is the meaning of the statement? That you are inserting one harmonic mean between A and B. What do you mean by that? It means that these three numbers are in Hp, right? Very good rationale. So we have already seen this, that H1 is going to be two AV by A plus B, correct? Now I would write this in a slightly fancy way. I will write this as, okay? And correct me if I'm wrong, two by one by A plus one by B, okay? Why I've written it like this? Why don't I leave it as two AB by A plus B? Because I want you to appreciate one pattern. So one pattern is going to be evolving from this, which I will be discussing subsequently. So if H1 is the harmonic mean between two non-zero numbers A and B, H1 is given by two divided by reciprocal of A plus, reciprocal of B, okay? So this was a case when I was inserting one harmonic mean between A and B. Let us extend this. If between A and B, I'm inserting two harmonic means, H1 and H2, okay? Please write down H1 and H2 in a similar format as what I have written for the previous case. Give me a response on the chat box. So what is H1 and what is H2 in terms of A and B and write your answer in the same format as what I have written over here. So the meaning is very clear to all of us that if these four quantities, if I'm inserting H1 and H2 as the harmonic mean between A and B, these four quantities are in HP, correct? So write down H1 and H2 in terms of A and B. See, you don't have to go out of the way to do anything. See, all you need to do is take that, these three quantities, these four quantities are in AP, okay? And now in your mind, think as if between one by A and one by B, you are inserting two arithmetic means, right? So we have already learned, please recall. I would write it at the side over here. Please recall this fact. Please recall that when they were two arithmetic means between two quantities, let's say A1 and A2 are two arithmetic means between, yes, so let's say these are in AP. Then what was A1? A1 was two X plus Y by three, isn't it? And what was A2? A2 was X plus two Y by three, okay? Same way, start treating your X as one by A, small A and start treating your one by H1 as A1. So one by H1 is equal to two X plus Y by two plus one, which is three. That means H1 is equal to three divided by two by A plus one by B. Similarly, one by H2 will be one by A plus two into one by B by three, which means H2 will be three divided by one by A plus two by B. Please note down the pattern. I would request everybody to pay attention to the pattern. What is the pattern? If you're able to identify the pattern, you will be able to write any number of harmonic means between two non-zero quantities. Has anybody seen the pattern? Has anybody appreciated what is the pattern going on? Just tell me by writing a me on the chat box. Who has identified a pattern? Okay, Sethu has identified a pattern. Chalo, we'll do one more just to see whether your identification is correct or not. Meanwhile, let me erase this off because this was just for your explanation purpose. Now, this time I'm inserting three harmonic means between A and B. So let's take another case. So between A and B, I'm inserting H1, H2 and H3. Tell me what is H1, tell me what is H2, tell me what is H3 in terms of A and B. Please write down your response for at least H1. I'm not expecting you to write everything down. See, look at the pattern. Pattern is self-evident. One harmonic mean, it is two. Okay, two harmonic mean, the numerator was three. Okay, so it's always one more than number of harmonic means. Okay. Now in one harmonic mean, you have one by one by B. Here you have to insert two harmonic means, so it is two by A plus one by B then one by A plus two by B, okay? So this is all basically hinting towards the fact that if I have to insert three harmonic means, the first one will be four by three by A plus one by B. Right, Vaster, correct her now. Second one will be four by two by A plus two by B and the third one will be four by one by A plus three by B. Is it not? Is it not? Yes or no? So can we generalize it now? So I think we have seen the pattern. So once you've seen the pattern, let's generalize it. So between two quantities, between two non-zero quantities A and B, if you're inserting N harmonic means H one, S two, S three, da da da da, till H one, okay? So what is your H one? H one will be N plus one by N by A plus one by B. H two will be N plus one by N minus one by A plus two by B. S three will be N plus one by N minus two by A plus three by B. And if let's say I go to any keyth harmonic mean, it will be N plus one by N minus K plus one by A plus K by B and the last one will be, the last harmonic mean will be N plus one by one by A plus N by B. Is it fine? Any questions, any concerns? Good enough? All right. So let's take a small question based on the same. First question that I would like you to solve, if H one, H two, H three till H N are N harmonic means between two non-zero quantities A and B, okay? Then, then, then find the relation between some of reciprocals of these harmonic means, the harmonic mean of A and B. So how are these two related? So how are these two related? So how are these two related? So how are these two related? That means if you do some of reciprocals of these harmonic mean, is it related to the harmonic mean of A and B? If yes, please find out that relation. See reciprocals of one by H I, what is the meaning of it? Basically you are trying to reciprocate all the harmonic means, okay? So the first one, if you reciprocate, correct me if I'm wrong, you'll end up getting N by A plus one by B by N plus one, isn't it? Second one will be N minus one by A plus two by B by N plus one. Third one will be N minus two by A plus three by B by N plus one and this goes on, this goes on till I reach, till I reach one by A plus N by B by N plus one, okay? Now take N plus one common and please be aware that if I take one by A out, it will be N plus N minus one all the way till I reach one. Similarly, if I take one by B common, it will be one plus two all the way till I reach it, okay? In short, it'll give you one by A, N into N plus one by two and one by B again into N into N plus one by two, divide throughout with N plus one. So if you divide through with N plus one, it'll give you N by two, N by two into one by A plus one by B, isn't it? Okay, can I write it like this? All of you please pay attention. Can I write it like this? N divided by two AB by A plus B, okay? So what does it become actually? It becomes N divided by the harmonic mean of A and B. Okay? So this is something which is a property which I would like you to take away here. So note down as an important property, some of the reciprocal of the harmonic means will be giving you N divided by the harmonic mean of A and B. Is it fine? Any questions? Any concerns? Okay, all good? Great. So we'll take one more question. Okay, let's take this one because this is one of the commonly asked questions in competitive exams as well. Find N if this term be the harmonic mean between A and B. If you remember, we had a similar set of questions for AP, sorry, for AM and GM as well. So now the same question has now been trained for HM. So find the value of N for which this term is the harmonic mean between A and B. Yes. Any response? Anyone? See, they're saying that this term is the harmonic mean between A and B. That means this is to AB by A plus B. Am I right? Oh, you want some time. Okay, fine, fine. Take your time. Not in a hurry. Should I now proceed? Okay. So let's take a cross multiplication here. So it becomes to AB by A to the power N plus B to the power N. One expansion, this gives you A to the power N plus 2 plus AB to the power N plus 1 plus B or A, A to the power N plus 1B plus B to the power N plus 2. And this will give you negative, sorry, 2A to the power N plus 1B and 2AB to the power N plus 1. Okay. Out of this, I think this gets cancelled with one of the other terms over here. And this gets cancelled with one of the terms over here. All right. That's absolutely right, Setu. So let's bring it to the other side. So it becomes A to the power N plus 2. I'm bringing this term to the other side and I'm bringing this term to the other side. Okay. So minus A to the power N plus 1B and AB to the power N plus 1 minus B to the power N plus 2. Okay. Take A to the power N plus 1 common. This will give you A minus B. Here you take B to the power N plus 1 common. This will give you A minus B. Now, since A and B are not equal to each other, since A and B are not equal to each other, you may cancel the factor of A minus B from both the places, which means A to the power N plus 1 is B to the power N plus 1, which means A by B to the power N plus 1 is equal to 1. Again, this implies two things. Either A by B is a 1, which means A is equal to B, which is not possible. Or the power itself is a 0, that means N is a negative 1. Okay. So this is the value of N. So here is a quick summary of this type of a question. So if you want A to the power N plus 1 B to the power N plus 1 upon A to the power N plus B to the power N to be the arithmetic mean of A and B, where A and B are not equal to each other. Remember, the value of N used to be, in this case, a 0. Okay. But the same thing if you want it to be the geometric mean of A and B, then N value used to be minus half. I don't know how many of you remember this. Okay. And now in the present question that we have taken up, if you want this quantity to be the harmonic mean of A and B, then your N value would be a negative 1. Okay. In short, these terms themselves are in AP. Isn't it? Isn't it? So 0 minus half minus 1, they themselves are in AP. Is it fine? Any questions? Any concerns? Anybody has? Good enough? All right. So now we are going to do a, you know, a very important concept, which is your AM, GM, HM properties. Okay. So now that we know arithmetic mean, geometric mean and harmonic mean, we are going to talk about some of the important and interesting properties of AM, GM and HM. Okay. So first of all, let me define what I mean to say AM over there. AM is basically arithmetic mean of, so let's say AM of N positive quantities. So please note that they are all positive quantities or they are all positive real numbers. Okay. Is given by, we have already seen this result. I'm just repeating it once again for you. Okay. Similarly, GM of these positive numbers is given by their product raised to the power of 1 by N or Nth root of their product and HM of these positive numbers we have already seen in today's session itself. It's N upon some of the reciprocals of these quantities. Okay. Now the very first property I would like to discuss over here, AM will always be greater than equal to GM and GM will always be greater than equal to HM. I think partly I had discussed AM greater than GM in our previous class, but now that I have introduced HM also to you, we'll be discussing it. But note down this is only for positive quantities. Okay. Please note you cannot apply this AM, GM, HM inequality like this for quantities which are negative or let's say even if some of them are zero, you cannot apply it. It's only applicable to positive quantities. However, this can be applied to even quantities where some of them are zero, but please note you cannot apply AM, GM inequality for quantities which are not positive. Okay. So only for positive real numbers, you can apply this inequality. Now I will not be proving this for N quantities. I'll be proving it for two quantities. Maybe let's say A1, A and B, let's say. So if I take two quantities A and B which are positive real numbers, I've already discussed with you that AM will be greater than GM. I'm doing the proof once again for you all. So we all know that this is true. Okay. So when I expand it, I'll end up getting something like this, which means A plus B will be greater than equal to two under root AB, which means A plus B by two will be greater than equal to under root AB, isn't it? And this is nothing but trying to say AM is greater than equal to GM. Now what I'm going to do, I'm going to use the same result a little further. So A plus B by two is greater than under root AB. So multiply it throughout with a two. Divide by A plus B. Okay. Remember, A and B are both positive. So dividing by A plus B doesn't change our inequality. So multiply throughout with AB. If you multiply throughout with AB, even that will not change the inequality. Right. And what do you see on your screen towards the right side? And is that I've written GM greater than equal to AM. Oh, sorry. GM greater than equal to HM. Okay. So from the transitive relation, I can say AM will be greater than equal to GM will be greater than equal to HM. Now when will the inequality hold true? So note down equality holds. That means they will all be equal when the two quantities themselves are equal to each other. That means in general, for AM of positive quantities to be equal to the GM of those same quantities should be same as the HM of those same quantities. All the quantities must be equal to each other. Okay. So I'll write it down over here. Equality holds. Holds. When you're a1, a2, a3, et cetera, they are all equal to each other. Is it fine? Any questions? Any questions? Situ, does it answer your question? All right. All right. Great. Now, next thing that you should all note down. If A and B are two positive quantities. Then the arithmetic mean of A and B, geometric mean of A and B, and harmonic mean of A and B will be in geometric progression. I don't know how many of you figured this out. So for two quantities A and B, their arithmetic mean, geometric mean, and harmonic mean themselves are in geometric progression. Okay. Let's figure it out how. The arithmetic mean is A plus B by two. That's obvious. Geometric mean would be under root of A, A, B. That's also obvious. And harmonic mean of A and B will be to A, B by A plus B, right? Now just do this activity. GM square. What do you get? A, B. And just do AM into HM. What do you get? You get A plus B by two into two AB by A plus B, isn't it? A plus B, A plus B goes off, two and two goes off. Didn't you get AB? So what does it mean? It means GM square is AM into HM, right? So clearly implying that AM, GM, and HM of two quantities A and B, are in geometric progression. Okay. Now many people ask this question, sir. Can it be true for, even if you have more than two quantities, let's say if there was ABC or positive numbers, can I say that AM, GM, and HM will be in geometric progression? What do you think? If I take any three quantities, or more than three quantities maybe, can you say that the arithmetic mean of those positive quantities, the geometric mean of those positive quantities and the harmonic mean of those positive quantities will be in geometric progression? Can you say that? Can you say that? Can you say that? Abna was saying, no. Okay. Anybody else? So in general, it is not going to be true. Okay. So please note that for two quantities it works, but it will not work for more than two quantities. Okay. For example, if I say, let me take one, two, three itself. Okay. Simple calculation for us. So these are all positive quantities. Okay. So let's find out their arithmetic mean. So arithmetic mean will be what? One plus two plus three by three. That's nothing but a two. Okay. What does geometric mean? Geometric mean will be cube root of one into two into three, which is actually cube root of six. Okay. And what does the harmonic mean? The harmonic mean will be three divided by reciprocal of these quantities. Okay. So if I'm not mistaken, this is going to be three divided by one plus five by six, which is 18 divided by 11. Right. I do not think so that six to the power two by three is equal to two into 18 by 11. Okay. That's not going to work out. Okay. So yes, in general, we cannot say this. Okay. So AM, GM and HM of more than two positive quantities or two positive numbers may not be in GP. Okay. But if the quantities themselves are in geometric progression, then yes, this property is going to work out. Let's see how, let's see how. So I think everybody has noted this down. Any questions anybody has to let me know. So now I'm going to show you one more facet of this particular property. Can I go to the next slide? Okay. So if a one, a two, a three, I'm just taking n numbers. They themselves are. So these are positive quantities and they themselves are in GP. Okay. Then the arithmetic mean, the geometric mean, and the harmonic mean of these quantities of these quantities will be in GP. Okay. Are in GP. Let's see how. So for the case of, for the sake of convenience, what I'm going to do is I'm going to take the quantities a one as a, a two as AR, a three as AR square, because I'm claiming that these quantities are in GP. Okay. So what should be the arithmetic mean arithmetic mean will be this. Okay. And we already know the, we already know the formula for some of n terms. So what will, what will this become r to the power n minus one by r minus one. So it'll be n r minus one. Correct. What will be the geometric mean, geometric mean will be the product of all these quantities raised to the power of one by n. So a to the power n, and this will be, get me if I'm wrong, this will be starting from r to the power one till you reach n minus one. So this will be the power. Okay. In short, this is going to become a r to the power n minus one by two. Correct. And what will be the harmonic mean, harmonic mean would be n divided by reciprocal of these quantities. Okay. Now what are the sum of the denominator over here if you if you see this very closely this, the denominator terms themselves are in GP. Okay. With first term as one by a common ratio as one by r. Okay. And number of terms are n of course, divided by one minus one by r. Let's simplify this a bit. Let's simplify this a bit. So this is going to give you, if I'm not mistaken, n divided by less, let's take r to the power n out and give you r to the power n minus one. Okay. Correct. And here I will get r minus one by r. Okay. So r minus one by r. One by r. So r will come over here and make it n minus one. Okay. So in short, in short, this will look like this n r minus one upon r to the power n minus one into a r to the power n minus one. Is it fine? Any questions? Any questions with respect to this? Anything that I missed out, do let me know. If you feel that I've missed out any terms, do please highlight. Okay. Now, if you multiply a m into g s or a m into h m. What do I get? So a r to the power n minus one by n r minus one. So if you multiply it with this term, which is n r minus one a r to the power n minus one by r to the power n minus one, you'll realize that this term gets canceled with this. This whole term gets canceled with this. So what does it give us? It gives us a square r to the power n minus one. And if you square your geometric mean, if you square your geometric mean means you square this term, that also gives you the same expression that is a square r to the power n minus one. So basically these two are equal. So what does it imply? It implies that, yes, g m square is equal to a m into h m, which means in this case, a m, g m, h m, r in g p. So here is a conclusion. If there are two terms, two positive terms, then a m, g m, h m will definitely be in a geometric progression. But if there are more than two terms, then a m, g m, h m will be in a geometric progression only when the terms themselves are in geometric progression. Of course, positive terms and they must be in a geometric progression. Are you getting this point? This particular concept has been asked in few competitive exams. And many people are not aware of it. Is this fine? Any questions, any concerns? OK. Next property, I think property. Can somebody tell me the property number which I'm taking? Sorry, I forgot the count of the property number. The first one was this second one, third one. Now it is third one. OK, thanks. A is the arithmetic mean of A, B, C. G is the geometric mean of A, B, C. And h is the harmonic mean of A, B, C. Then A, B, C are roots of the following equation. A, B, C are roots of the following equation. Now, let us prove this. It's very simple. Oh, it's a fourth one. No issues, no issues. It's just a small change I have to make. Yeah. So this is a property which comes very commonly in competitive exams as well. So if capital A that you see in your cubic equation over here is the arithmetic mean of A, B, C. And capital G is the geometric mean of A, B, C. And capital H is the harmonic mean of A, B, C. Then A, B, C themselves are root of this cubic equation. OK. Let's prove it. It's very simple. And I would request you to write a done on the chat box once you are done with the proof. Done, anybody? Excellent. Excellent enough. See, if A, B, C are the roots of any equation, any cubic equation, how do you formulate a cubic equation from the fact that you know the roots? So we all know that this is the cubic equation whose roots will be A, B, C. Isn't it? A, B, B, C, A, C, X minus A, B, C equal to 0. Now, I have to write the very same equation in terms of capital A, capital G, and capital H. So we already know that capital A is the arithmetic mean of A, B, C. A plus B plus C is equal to 3A. So let's call it as 1. Second thing that we know that capital G is the geometric mean. So can I say G cube is equal to A, B, C? Or A, B, C is equal to G cube. Let me call this as second one. And we also know that capital H is the harmonic mean of A, B, C. OK. So when you write this, you basically write something like this. H is equal to 3A, BC by AB, BC, AC. OK. AB, BC, AC. And this is nothing but H is equal to 3G cube by AB, BC, AC. In short, AB, BC, AC will be equal to 3G cube by H. OK. Let me call this as the third one. Now, all I need to do is filling these values with our A, H, and G terms. So this is going to become X cube minus 3AX square. This term is going to become plus 3G cube by H, X. And the last term is minus G cube equal to 0. OK. And this is what we wanted to prove, hence proved. Now, many students asked me, so do we need to remember these equations? No, not required. But it may be a possible question in your exam. OK. So now we are going to take up problems which are based on AM, GM, HM inequality. I hope you have all noted this down. Roots of any cubic equation. Why would it be roots of any cubic equation? It doesn't help you to find the roots. It basically tells you that if you know the roots, the equation can also be written like this. The cubic equation whose roots are those quantities can be generated like this. That is what it tells you. It doesn't help you to solve it. Because you are only aware of the terms. Or if you are only aware of these expressions, it doesn't help you to solve the problem. It just tells you that if ABC are the roots of this cubic equation, that cubic equation coefficients will be related to the arithmetic mean, geometric mean, harmonic mean in this manner. See, you want to make a cubic equation whose roots are ABC, OK? So this is the cubic equation whose roots are ABC. You know this, right? I'm not aware of this also. So in the same thing, we have to put our A plus B plus C, AB, BC, CA, ABC, etc. in terms of arithmetic mean, geometric mean, and harmonic mean. That is what I have done over here. So this term A plus B plus C is nothing but 3A. So I have replaced 3A over here. This term ABC is G cubed. So I have replaced G cubed over there. And from the harmonic mean, I got an expression for AB, BC, CA, which is this one, the one I have written over here. So that also I replaced over there. That's what I wanted to prove. So what type of questions can come here in this? They will say, if arithmetic mean of ABC is A, A, A, geometric mean is capital G, harmonic mean is capital S, then ABC are roots of which of the following equation. And this equation will be sitting in one of your options. Next, we'll take problems on AM, GM, HM inequality. AM, GM, HM inequality. So please note that this inequality is one of the very commonly asked inequalities. I'm sure you would have experienced it in your trigonometry chapter itself. Some of you pointed out that this is how they have done this problem. I don't understand how have they done it. So they had used AM, GM, HM inequality in many of the questions. Some of you had pointed that out to me. So this is a widely used type of inequality that you will find being asked in many comparative exams. So let's take a few questions just to get an idea of the process. So the first question that I would like you to solve here is this. Given the greatest value of XYZ for positive values of XYZ, subject to the condition that YZ plus ZX plus XY is equal to 12th. Now, can you see the questions? Have you seen the question in half pieces? That's a problem with the Zoom sharing. Even I was seeing it as a half question, but I was waiting for somebody to point it out. Some bug, some bug in the system. See, I hope the question is clear to all of you. The question is there are three quantities, XYZ, they're all positive quantities. You have to find the greatest value of XYZ, subject to the fact that YZ plus ZX plus XY is equal to 12th. Anybody? Okay, Siddharth. Let me solve this one question maybe from this question, you get an idea how to solve it. See, if XYZ all are positive quantities, can I say YZ, ZX, XY, they will also be positive quantities. So can I do one thing? If they are positive quantities, I can say arithmetic mean is greater than equal to the geometric mean. Okay, Arnav. Arraf. Okay, Prisham. So arithmetic mean that means YZ plus ZX plus XY by three will be greater than equal to YZ into ZX into XY to the power of one third. In short, what I want to say is that 12 by three will be greater than equal to, now remember, give you X square, Y square, Z square to the power of one third. So this is four greater than. Now just take a, just take a power of two out from each of these terms and write it as to the power of two by three. Okay, now let's raise both sides to the power of three by two. So this will be greater than XYZ. In short, XYZ would be lesser than equal to eight. So if XYZ is lesser than equal to eight, it clearly implies the greatest value of XYZ is equal to eight. Okay, so well done. I think Arraf gave this answer. Is it fine? Any questions? Any questions with respect to this? You could use HM, GM HM inequality also, right? See, if you use that, you can say XYZ to the power of one third would be greater than equal to three divided by one by X, one by Y, one by Z. Okay, so XYZ to the power of one third would be greater than three XYZ by XYZ ZX. Okay, in short, you're trying to say that. Correct me if I'm wrong. XYZ ZX by three is greater than equal to three times, sorry, three is already taken care of. XYZ to the power of two third. Isn't this the same as the step? That will give you the same answer. Okay, so whether you take AMGM or GM HM, result would be the same. So this is the second way of solving the same question. Is it fine? Kardik, clear? Okay, can we take another one? Let's take another one. Let's take this one. Find the greatest value of, find the greatest value of XQY to the power of four, subject to the fact that two X plus three Y is seven and X and Y are greater than equal to zero quantities. But this question is an interesting one and I think I should solve this one for you. See here, this term gives us a clue. Okay, this term basically involves X, X, X, Y, Y, Y, Y. So there are three X's and there are four Y's which have been multiplied to get XQY to the power four. Okay. Now here, the catch is, if you want to make this term appear in the geometric mean, normally these kind of terms will appear when you are taking a geometric mean, then you should be taking geometric mean of seven terms, three of which should be having X and four of which should be having Y in it, isn't it? So this is what I'm going to do. I'm going to break this two X as two X by three, two X by three, two X by three. Okay. And I'm going to break this three Y as three Y by four, three Y by four, three Y by four, three Y by four. Okay. Now see what I'm going to do. So I'm going to treat these seven terms, one, two, three, four, five, six, seven terms. Of course they're all greater than equal to zero. So I will say the arithmetic mean of these seven terms, please note, since there are seven terms, you have to divide by seven. This should be greater than the geometric mean of these seven terms. So two X by three, two X by three, two X by three, three Y by four, three Y by four, three Y by four, three Y by four, four less to the power of one by seven. Okay. Now I think many of the threes will get canceled off. So let me do the honors here. This, this, this, this, this, this goes off. Okay. And I think two, two will go off from here. Okay. So I think I'll be left with correct me if I'm wrong. I'll be left with X cube. In fact, this will go by a factor of two. So I'll be left with X cube into Y to the power four times three by 32. Is it fine? I hope I've not missed out anything. Holes to the power of one by seven. Okay. Now on the left-hand side, this is actually two X plus three Y only, right? I've just written it as, you know, a fancy way for you to understand that I was basically trying to do AM greater than equal to GM on these numbers. So X plus three Y itself is a seven. So seven by seven will be a one. So one will be greater than three by 32. In fact, you can raise both sides to the power of seven. It doesn't make any difference. So in short, you get 32 by three greater than XQ, Y to the power four. What does it mean? It means that the greatest value, the greatest value of XQ, Y to the power four will be 32 by three. Is this fine? Any questions? Any questions? Any concerns? I'm sure you would like to solve one more of, you know, this kind. So I'll just give you one more and then we'll start the discussion for series. See, that's what I said. I basically looked upon this term XQ, Y4. Okay. XQ, Y4 will only come as a geometric progression. When you are multiplying three such terms, which are having X and four such terms, which are having Y. So that two X plus three Y itself, I broke it up in that fashion, right? So automatically this term XQ, Y to the power four appeared. Don't worry. I think this example would give you a new learning and you will be able to apply this on our next question. So next question is, if X, Y are positive quantities, both are positive quantities and X, X square YQ is a six. Let me write this in a proper way. Then find the least value of, find the least value of three X plus four Y. Very good, Situ. Very good. How about others? That much, I don't know. Are you sure? Check your working once again. Same approach. Since there are two X's and three Y's involved, what I'm going to do, I'm going to break this three X as very good. That's correct. I'm going to break this up as three X by two, three X by two, and I'm going to break four Y as four Y by three, four Y by three, four Y by three. Okay. Now there are five terms in all together. So what I'm going to do, I'm going to apply AM greater than equal to GM these five terms. Okay. So there are five terms in all. So three X by two plus three X by two plus four Y by three plus four Y by three plus four Y by three. Absolutely. Very good. This term would be greater than the product of these five terms raised to the power of one by five. Okay. Now this is as we were saying three X plus four Y by five is greater than equal to. Now here also a few terms would stand to cancel. So it becomes 16 by three X square Y cube to the power of one by five. Now X square Y cube is already given to us as a six. That means three X plus four Y by five will be greater than 16 by three times six four raised to the power one by five. So this is 32 raised to the power one by five. That's nothing but a two. That means three X plus four Y is greater than equal to 10. That means the least value. See it is greater than always later than equal to 10. So least value is 10. Is it fine. Any questions. So with this we are now going to cover up the series part of the chapter. So series as I already discussed in brief with you series is nothing but some of terms of a sequence. Right. So what are we going to learn under this particular sub topic. We are going to learn how to find some of certain terms in a sequence. Okay. Of course AP sequence GP sequence you already know. So we are going to extend to many more other series. Okay. We'll see. So I would try to I would like to discuss first of all special series with you. Okay. So there are three series. In fact there are some series which we call as special series. And why they call special series because they are the building blocks for us to find out the sum of some slightly tweaked version of the series. Okay. So I would first start with special series. Let me write it in white. So under special series we all need to learn what is the sum of all natural numbers from one to M. I'm sure you would say sir we already know this. Okay. But still I'm repeating it. And as a matter of symbol I will write it like this. Summation of R from R equal to one to L. Okay. This result is given to is already known to us from our you know arithmetic progression. Some which is N N plus one by two. Now I will derive this result in a slightly you know different manner. What time is break break happens at around six six fifteen. Now it's only five. Five thirty. Ah. Both classes. They are not being a part of so many classes. Last two classes you are not there. Right. Okay. So I will prove this. I will prove this in a slightly weird way which probably would be a new way to prove this. So this result I'm going to prove it in a slightly weird fashion. I would not say weird fashion. New fashion I would say. I would like to first ask you this question. What is K plus one whole square minus K square? I mean as an identity if you have to complete it. What do you get? Write it down on your chat box. Two K plus one. Correct. Okay. Now in this expression start putting K value as one. Two. Three. All the way till N. Okay. So in the very same expression I'm not going to change anything out of I'm not going to change anything out of I'm not going to change anything out of I'm not going to change anything. I'm not going to change anything out of I'm not going to simplify or do anything. I'm just going to put K as one. Two. Three. Four. Five. Till I reach in in the very same expression. So when I put a one I get two square minus one square is two into one plus one. Okay. By the way, don't treat this as two point one. It is two into one. Okay. If I put it to I get three square minus two square as two into two plus one. If I put a three I get four square minus three square as two into three plus one. Okay. And carry on till you reach the last one which is N plus one whole square minus N square. This will be two into N plus one. Okay. Add it. When you add it you would see surprisingly all these alternate terms will start getting cancelled off. Okay. Leaving behind. Leaving behind only these two N plus one whole square and minus one square. So this guy will survive. And this guy will survive. Okay. On the right hand side if you sum up the entire terms coming in this particular column. So it will be two times one plus two plus three till N which I want to find out. Okay. For the sake of convenience let me call this as let's say S. Okay. So I can call this as a two S. Anyways I'll write it down in the next step. And if you sum up this one one once you'll get an N. So you end up getting N square plus two N is equal to two S plus N. Which means N square plus N is equal to two S. So S is nothing but N square plus N by two. Which is nothing but N into N plus one by two. Interesting right. So I didn't use an AP sum which you already knew. So I thought I would do it in a slightly different way. But what is important is this process because I'm going to use this process to prove few more things. So notice down if you want to. If you're only familiar then it's fine. Is it okay? Any questions? Okay. So with this I'm going to move on to the second type of special series. Let's take that up. Next is sum of squares of natural numbers. Again this result is also known to many of you. Symbol wise we write it summation R square from R equal to one till N. Okay. So this result is N N plus one to N plus one by six. Okay. Again, how do you derive it? I'll give you again the proof for this derivation. By the way, this is not an AP. Okay. So AP formulas are not going to work. This is not a GP. GP formulas are not going to work. Right. So this series is a special one and it is going to be very helpful to know this result. Let's derive it. So if you remember for proving the sum of natural numbers, I began with an expression which was K plus one square minus K square. Right. So here any guesses what I'm going to begin with? Any guesses? Before I complete it, can somebody write it down? I write our left. So K plus one cube minus K cube. So we all know that K plus one cube minus K cube will give you three K square plus three K plus one. Okay. In this identity, start putting your K value as one, two, three, and so on till you reach N. Okay. So when you put a one, you get two cube minus one cube as three into one square. And before that, I'll let me call this as an S. When you put a two, I get a three cube minus two cube, three into two square, three into two plus one. When I put a three, I get four cube minus three cube as three into three square plus three into three plus one. And I reach N plus one cube minus N cube as three into N square plus three N plus one. Okay. Just add the both sides of the equal to side. So the left hand side, just like we had it in our previous case, will lead to cancellation of many of the terms leaving behind only N plus one cube and one cube. Whereas on the right hand side, this will give you three times one square, two square, three square to N square, which is your S. And three times one plus two plus three till N, which is N, N plus one by two. We have already figured this result out. And one plus one plus one plus one plus one plus N times. So that is a, anyways, an N. Okay. Let's try to make S the subject of the formula. So let's try to cancel off some terms from here. This will be N cube, three N square plus three N. This is three S. This will be three N square by two, three N by two plus N, which is going to be five N by two. Correct me if I'm wrong. So far, so good. Any questions? Any questions? Actually, one more thing before I forget. May not sound very good to you all. December 1st to December 15th. Your seniors are having their board exam. Okay. Unless until there is a postponement of the exam, during that duration, you will be having some extra maths classes. Okay. So normally one class you have per week, maybe during that duration only, I'm just talking about the last first two weeks of December, you may be having extra maths classes. Okay. So just to expedite the process of completion of our topics. Don't worry. All the reminders, et cetera will be duly sent to you. I'll keep you updated one week before we start doing it. I'm talking about December. So full November is there as of now. Don't worry. It depends upon the faculty member for maths. I am the decision maker. So I know I need more classes. I'll take it. If he doesn't feel he needs more classes, he will inform you accordingly. Same with promises. Good morning. Normally I keep pinging people saying they're attentive or not. I caught you. Okay. This is going to give you this. If I'm not mistaken, this is, yeah. So from here I can say, if I'm wrong, three s is equal to two n cube plus three n square. Plus n. All divided by two. In fact, I can take a n common and it'll be two n square. That's three n plus one by six. Okay. This term itself is factorizable. And I'm just going to write down the factors here and plus two n plus one into n plus one by six. And there you go. This is the result that we had already it. Okay. Is this fine? So these all results have to be kept in mind because they are going to help us solve, you know, many complicated series, which we are going to see in some time. Okay. Similarly, I'm going to give you one more result, which you should keep in your mind summation of cubes of national numbers from one to n. And I'm leaving the proof of this up to you. Okay. This is called summation R cube. So I'm leaving the proof of this up to you and the summation result is this. Okay. So please prove it for homework. Proof the above result. Okay. And you know the process. You have to start with K plus one to the power four minus K to the power. Sir, could you code? Good. Yeah. Okay. Done. Now many people ask me, sir, why don't you give the result for fourth power of all national numbers? One to the power four, two to the power. See, I can always derive it, but up to that power we don't require. We only require the three special series, which I've given over here. And let us try to see what type of questions will be asked and what is the process where these special series will be helpful. Is it done? Can I move on to the next slide? So please note this result down also. Okay. So now we are going to see where is this special series used. So special series is utilized in a method which we call as Sigma method of summation. In order to explain this process, in order to explain this method, I will take a simple question. Okay. Let me pick up a question. The question is very simple. I want to sum this series up till end terms. Okay. So what is the series? One into two. By the way, don't read it as 1.2. No, it's not 1.2, 2.3, 3.4. It's one into two, two into three, three into four, four into five, up till end terms. Okay. So how do I sum this particular series? Please note it is not in AP. It is not in GP. It is not the sum of national numbers. It is not the square of sum of national numbers, et cetera, et cetera. Okay. So the process is very simple. Please listen to the process. Process is very important. So first thing is step number one. We write down the R-th term. What do we do here? We write down the R-th term of this particular series. Can somebody write it down? What is the R-th term? Oh, no problem, Ashraf. You can watch the recording where you missed. Please focus on whatever is going right now. Write down the R-th term of the series. All right. R-th term of the series is R, R plus one. Correct. Now what do I have to do? We have to find sum till end terms of the series. Sum till end terms of the series is nothing but sigma TR, R from one to N. Okay. Now here in place of TR, we're going to write R, R plus one. Okay. Let's break this down as R square plus R and I have to do a sigma of this. Okay. Let's write it down as two separate sigmas. Okay. So please note that in sigma, you can use this fact that you can sum up both of them separately. Okay. In fact, if you have a constant multiplied to them, then the constant can be pulled common out also. Okay. Now once you have achieved this, the third step is just recalling your special series. Have we done what is the sum of all squares of natural numbers from one to N? In short, have we done what is summation R square from one to N? Right. Yes, we have done it. So this result, you put it down over it. Okay. That is why it is important to remember those results. Have we done some of all national numbers from one to N? Yes. Please put the result down. Okay. In fact, you can do a further simplification, but I mean, I would consider the process to be over at this stage. After that, you can just do a simplification that is up to you. Okay. Or up to the options given to you. So you can take N into N plus one by two common. You'll have to N plus one by three plus one. That's makes it N into N plus one by two into two N plus two by three. That makes it N into N plus one into N plus two by three. Okay. So this is the answer to this particular question. Now, whenever you are solving a summation question, always verify whether your result is correct or not. So how do we verify? Very simple. Just choose N value as two. Let's say this is SN. Okay. So N basically means how many terms you are summing up. So N basically, if you pick up as a two, I should get the answer as two plus six. That is eight. So does your answer give you an eight when you put NS two? Let's check. Does this work out one into two, two into three? Does it give you eight value? Let's check. So as for this result, my answer should be two into three into four by three. Yes, very much. It gives me an eight. Right. Okay. Okay. So now you realize what is sigma method of summation? What is the process involved? And where did those special series help you out in finding the sum of these kinds of a series? In any one of your schools, was this process or was this method done for series? Or was it completely washed off the syllabus? Deleted. Is it fine? Okay. Would you like to take more questions now? Process is clear to everybody. Right. Let's take this question. So when you're solving it, I don't want you to simplify unnecessarily. I mean, leave your answer in a raw form. No need to simplify. And remember the steps. What I told you. Right. You're on it. Correctly. Why I'm saying correctly. 90% people write it wrong in the initial part of their learning process. Okay. Step number one. What is the earth term? Now, see, it's term itself. Has a numerator which will go up till. RQ. And it has a denominator. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. RQ. It doesn't matter. RQ. How are you going to do the same? if I'm going to R cube, the summation there would be R R plus one by whole square. Now remember one cube to cube till N cube was N N plus one by two whole square. So the same N you put it as R here, that's it. Okay. And then the denominator, you actually have a arithmetic progression. So arithmetic progression, the sum is number of terms first term plus the last term, isn't it? I hope you all remember your formulas. So this is R square R plus one whole square by four and down in the denominator, you'll end up getting an R square. Check it out because this one and this one will get cancelled. This two and this two will get cancelled. And not only that this R square and R square will also get cancelled. So you end up getting R plus one the whole square by four as your Rth term. Please verify it. If I put R as two, do I get nine by four? Check. If I put R as one, do you get one? Check. Yes, you're getting. See at every nook and corner, you should keep checking your result, whether it is actually correct or not. Don't be complacent at all. Whatever I've done is correct only. No. Keep checking it. Even we teachers keep checking our results. Okay. Now first step is accomplished. Second step was some till end terms is nothing but summation dr R equal to one till end. Okay. So it's summation of R plus one whole square by four. Okay. Now there are two ways to solve it over here. I'll show you first very, I can say not so smart way. Okay. So many people solve it like this. They will write one fourth R plus one whole square and they will expand this term and write it like this R square plus two R plus one. Okay. R equal to one till end. And then they will split it up as summation R square from R equal to one till end to summation R R equal to one till end and summation one summation one is going to be in itself. Okay. Now see summation R square from one to end that result is already known to us. Right. That is why those special cities were done for you. Summation R is also known to us and into n plus one by two. Okay. So this becomes your answer. Of course, I have not simplified it. It can be further simplified. So this is one way to solve the question, but I feel this way is not a smart way. Okay. There's a smarter way to actually do this process. So this is one way to do it. I'll tell you another way to do this. First note this down. And if you have any questions, do ask. Another way to do the same thing is I'll act slightly smart over here. See, let me put a substitution over here. Let me put R plus one as a K. Okay. Now if I put R plus one as a K, please note that this term will be summation K square. But the only change that will happen here is now your summation will go from two to n plus one. Why? Because when R is one K is two and when R is n, K is n plus one. Correct. Now I'm going to write this further as summation from one to n plus one or K equal to one till K equal to n plus one K square minus put one square subtract it. That means subtract the one. So see instead of going from two to n plus one, I decided to go from one to n plus one and just subtracted the very first term that I would be getting, which is one square. Now this makes our life slightly easy because summation K square. Remember summation R square summation K square doesn't make any difference. It's just the name of the variable, which is changing. Okay. So summation R square from one to n result is known to us. What will be the summation of R square from one to n plus one? Then you'll say simple, you just change your n with n plus one. That's it. In other words, your result will become something like this. Is this understood? And trust me, both of them on simplification will be the same things. See, this is like saying, this is like saying you're doing two square, three square, four square up till n plus one square, right? They just funny. I want to find this. So what I did here, I added a one square to this and I subtracted a one square. This is what I have done over here. Understood. Why did you do that? Because there is no direct formula to sum up from two square. The formula is only from one square till whatever square you want to go to. What did they just mean? So in this expression, in this expression, you can basically write the formula of summation R square. But in that formula, you need to replace n with n plus one. Is it fine? Any questions? So now please note that if you sum till n square, your formula is n n plus one by six and n plus one to n plus one by six. But now note that you are summing till n plus one. In the very same formula, you just replace n with n plus one. That's what precisely I have done over here. Any questions? So with this, we now take a break. We'll be continuing with a method of difference on the other side of the break. Now, many times, you would be provided with some special type of series where the nth term is not very evident. Okay. But you will see some patterns in that particular listing of those terms. So let me take such example to illustrate what I want to say. So let's say I have a question like this. Let's take this question. Now, if you see this question, this question has been just asking you to find out the nth term of the series. It's not even asking you to find the sum till nth term. Maybe we can, we can find it out separately on our own. But here the challenge itself is the pattern is not so evident that we can figure out the nth term very easily, right? Because in order to find the sum using our sigma method of summation, the very first step was to get our rth term. And rth term will come from our nth term. Okay. And r doesn't make any difference in the name. So here the main challenge coming up for us is that we are not aware what is the rth term or nth term of the series. Okay. So let us try to address that problem first. Can anybody tell me, is there any pattern that you can see in the listing of these terms? One, four, 10, 20, 35, etc. I mean, they're not given a lot of terms, only five terms. From there only we have to figure out if at all there is a pattern hidden in it. I mean, I don't know. I'm not asking you the nth term or rth term here directly, but at least can you identify that there is a pattern hidden in these terms? If yes, what is the pattern? You can unmute yourself and talk if you want to. Anybody? No? Let me just list down the terms only. Okay. Dot, dot. Try to take the difference of these, you know, consecutive terms. So what is the difference here? Three difference is six, 10, 15. Okay. I can take the difference three, four, five. Okay. So here you realize that you are now getting an arithmetic progression. Isn't it? Yes or no? Okay. So when there is a scenario where you have been given a series or certain sequence, who is the difference of terms? Okay. By the way, let me tell you some, you know, nomenclature that we use over here. We call the terms themselves as delta zero T. Okay. Why delta zero? Delta stands for difference. Delta zero means no difference. I'm not taking any difference. That means that we are using the terms themselves as given in the question. This series is called delta one T. Sometimes you just call it as delta T. And this difference we call it as delta two T. Delta two T means you're taking difference of the difference. That is why we call it as delta two T. Okay. So in such kind of a question where you realize that the difference or the difference of the difference or the difference of the difference of the difference or the difference of the difference of the difference, et cetera, eventually comes out to be an arithmetic progression. Then how do we find out the nth term of that particular series that is number one? And how do we find out the sum of the nth term of that series? Okay. Of course, if you find the nth term, finding some becomes easy because we already know our special series. Okay. So in order to do this, there has been some observation which has been done by mathematicians in past. And I'll be sharing that observation with you. See, when, when the terms themselves are in a geometric progression, okay, sorry, arithmetic progression, we have already seen that when the terms are in arithmetic progression, your nth term is a linear expression in n, right? Right? Isn't it? So when the term themselves are in arithmetic progression, let's say I take any arithmetic progression, one, three, five, et cetera. Okay. Your nth term here is a linear function of n, isn't it? So in this case, it is going to be 2n minus one, right? But when the difference of the terms are in arithmetic progression, then your nth term will be a quadratic expression in n. Are you getting this point? And when the difference of the difference is an arithmetic progression, then your nth term would be a cubic expression in n and this trend continues on. Okay. So I will tell you a theory over here. Please note it down. Okay. If your delta zero t or delta, without any difference, if the terms are in AP, then remember your nth term is a linear expression in n. Okay. That we already know because in an AP, we get the nth term as a linear expression in n. Linear means the power of n is going to be one. Okay. But if you realize that your delta one t are in AP, that means initial terms are not in AP, but when you take a difference, the first difference that is delta one t, that is in AP, then your nth term would be a quadratic in n. Okay. Similarly, if the difference of the difference, that means your second difference are in AP, then your nth term would be a cubic expression in n, something like this. Okay. If your third difference is in AP, okay, then your nth term would be a bi-cordratic expression in n. That means the fourth degree expression in n. Of course, there is very, you know, bleak chances that you will be, you know, asked to go to this extent. But at least if you, you know, observe this pattern, you can solve a lot of questions very, very easily. Okay. Now I will not be going into the proof for this. Okay. The proof for this is not required. We'll be just talking about the application of this concept to solve this question. So here you realize that your delta two difference gives you terms which are in AP. So this means that your nth term would be a cubic expression in n, something like this. Right. And all I need to do is figure out what is this a, b, c, and d values, then my job is done. So I can basically get what is my nth term of the series, which is what the question is asking us. Okay. So how do I get a, b, c, and d? Very simple. Put n as one and equate it to the first term. So first term is four and n is one will give you a plus b plus c plus d. Put n as two, equate it to the second term. Put n as three, equate it to the third term. Put n as four, equate it to the fourth term. Okay. Now you must be wondering, are you, sir, it leads to four equations and four unknowns. See, they're very easy to solve. Keep taking the difference of the two consecutive ones. So take the difference of these two. Seven a plus three b plus c is three. Take the difference of these two. Nineteen a plus five b plus c is equal to six. Take the difference of these two. You get 37 a plus seven b plus c is equal to 10. Okay. Again, take the difference of these two. Twelve a plus two b is equal to three. Take the difference of these two. Eighteen a plus two b is equal to four. Again, take the difference of these two. You get six a is equal to one. So a will be one by six. Easy. In one shot, we got a. Okay. So if a is one by six, that means put it in the, any one of the two equations here through the last ones. So twelve a will be two plus two b equal to three. So b will be half. Okay. So a is six. Sorry, one by six. b is half. Okay. Then put it in any one of these three equations. Let's say I put it in the first one. So seven a, three b, three b will be three by two plus c is equal to three. So c will be equal to three by two minus seven by six. If I'm not mistaken, this is nine by six minus seven by six, which is one by three. So c is one by three. Okay. So c is one by three and if c is one by six, b is one by two. C is one by three and D is supposed to be found out. By the way, this is already a one. So D will be zero. Okay. So having got these values, you just have to plug in in this expression of TN and your job is done. Okay. So your answer for the nth term of this particular series is one by six then cube half n square one by three and that's your answer. Now don't get complacent. Never ever get complacent when you're solving series based question. Always do a quick check. Right. Put n value as two here and check whether the answer that you get from here matches with four because your second term is a four. Okay. So put n value as two. So when you put n value as two, this becomes eight by six. This becomes a two. This becomes two by three. So if I'm not mistaken, this is going to be two plus 12 by six. Right. That is actually a four. Yes. So that is verified. Okay. So that gives you the answer. Is this patently to everybody? Now I know what all questions will be coming in your mind. So what if the difference came out to be GP? Then what to do? Don't worry. I'll come to that part also in some time. Okay. Now this question was only asking you for the nth term, but let me extend this question to finding the sum of the series also. Okay. So for some of the series, you'll say, I can use my sigma method of summation, the ones which we discussed. So once you have got your nth term, which means you have got your rth term also, which is r cube by six, r square by two and r by three. Sum till n terms is nothing but sigma dr, that r from one to n. That is nothing but sigma of this expression, which is r cube by six, r square by two, r by three. Just break it up as three separate summations. So there will be one six summation r cube, half summation r square and again one third summation r. And by the way, you all are well versed with these formulas or you can just plug in the formulae and the result is there. So it's one, six nn plus one by two whole square, half nn plus one to n plus one by six and one third n into n plus one by two. Okay. Of course, simplification I'm not doing. Don't expect me to do simplification. Okay. So please try it out. Please do it on your own. Is it fine? Any questions? Okay. Now what if I make your life even more easier? I mean, you don't have to, I don't even do this. Will you be happy? Yes, sir. Very happy. Okay. So let me make your life even more easier. In fact, I will not make your life easier. There was one great scientist of all time. He has made our life easier. Sir Isaac Newton. So Newton has given a very interesting approach to solve problems of this type. Okay. And that approach is called the Newton's difference method or Newton's difference formula. So I'll be discussing that also with you. But before that, if you have any questions, any concerns, please do highlight any questions anywhere in this process. Einstein gave and its name the Newton's didn't get that. The jokes are like tangential to me. Okay. Can we go to the Newton's difference method? Okay. Same problem. I'll put it once again and I'll solve it by what Newton had basically given. Again, I'm not going to give up proof for this. Just understand the process. Okay. So what did Newton do? He basically first wrote down this series of course. Okay. He started taking the difference like the way we did it for the previous question. It took the difference of this as well. It took the difference of this as well. And it took the difference of this as well. Einstein. Sir Isaac Newton, I said Einstein gave this method. No, I never said Einstein. Did I say Einstein? I said it was given by one of the greatest scientists of all time, Sir Isaac Newton. I said Newton only. Get your ear checked. Okay. Now see, so what did Newton do? He basically said that if you have been given certain terms where if you take consecutive difference like this and eventually you start getting 0, 0, 0, 0, 0, 0, he said that in such cases, your nth term is, now listen to this very carefully, 1, this 1 into n minus 1 c 0. Now you must have understood that this is your binomial coefficient term, the ones which we were using in our bridge course and limit chapter also. And of course, he summoned upon this method when he was actually working on binomial theorem. So as you all know, Newton was the one who was responsible for binomial theorem also. Now see, this 1 into n minus 1 c 0, this 3, so basically I'm focusing only the first first terms of every difference, n minus 1 c 1, then this 3 into n minus 1 c 2, this 1 into n minus 1 c 3, that's it. This is the nth term of the sequence. Okay. By the way, I will write it in a more simplified way. I will not leave it in terms of this. By the way, I'm expecting that everybody is aware of ncr expansion. What is ncr expansion? n factorial, r factorial, n minus r factorial, correct? Everyone knows this. Anybody with idk? I don't know this. Okay. Now, so n minus 1 c 0 is a 1 only, n minus 1 c 1 is n minus 1, n minus 1 c 2 is n minus 1, n minus 2 by 2 factorial, 2 factorial is 2 only. And this is n minus 1 n minus 2 n minus 3 by 3 factorial, which is 6. Of course, this is in a very raw form. It is not simplified. But let us check whether actually Newton was correct. Let me put n value as a 4 and let's check whether I'm getting 20. In this, I'm putting n as 4. Let's check whether 4 gives me 20 or not. So 1 plus 3 into 3, so 3 into 3 into 2 by 2, 3 into 2, sorry, yeah, 3 into 2 into 1 by 6. Let's check. 1 plus 9 plus 9 plus 1. Yes, I'm getting 20, which is basically matching with the fourth term. So you don't have to do even the shortcut method. So this is the shorter shortcut. Okay. So thanks to Newton. Thanks to his, you know, tons of work. This was one of the small works he had done where you could find out the nth term. But again, it will only work when this particular aspect is satisfied. Okay, difference of difference of difference eventually starts giving you 0, 0, 0. And now to make you even happier, he also gave a formula to find the sum of n terms for such cases. So even that I can, I will give you now. Okay. So first note this down. So again, I'll repeat the process here. The process is he started taking the differences. Just focus on the very first term of your delta, you know, differences and start writing it as the first, this term n minus 1 c0, always n minus 1 c0, you have to start with. Okay. n minus 1 c0, then 3 into n minus 1 c1. Again, this 3 into n minus 1 c2. And again, 1 into n minus 1 c3. Don't worry, we'll take some problems also. Can you explain this expansion of n minus 1 c2? Oh, n minus 1 c2. n minus 1 c2 from here. It'll be n minus 1 factorial by 2 factorial, n minus 3 factorial. Am I right? Now n minus 1 factorial is n minus 1, n minus 2, n minus 3 factorial. n minus 3 factorial, n minus 3 factorial will get cancelled. So it'll be n minus 1, n minus 2 by 2 factorial, 2 factorial is 2 only. That's what I wrote here. A factorial property, everybody knows. No, I've done that factorial property also earlier. Okay. Sevedu convinced? Okay. So now Newton gave a method to find the, so please note this down, this is important. Okay. So he gave a formula to find the sum of n terms also, which is again the same number 1, that is this term, first term here, this fellow 1. And instead of n minus 1 c0, just add one more to the superscript and subscript. So it'll become nc1. Okay. Again, 3 into nc2. Again, 3 into nc3. Again, 1 into nc4. That's it. Okay. And I'll write down the expansion of this also in case you want it. Is that fine? We'll let's verify. We'll not become complacent. We'll verify. If I put n as 3, I should get 15. Put n as 3 here. Am I getting 15? Check. 3 plus 3 into 3 into 2 by 2. 3 into 3 into 2 into 1 by 6. This will anyways be a 0. So I'll get 3 plus 9. Oh yes, I'm getting 15. 25. So this formula is going to work. Is it fine? Any questions? Any concerns? Again, this is not mentioned in many books. Okay. This concept has been picked up from you know, higher algebra books like Bernard Schild, Holland Knight, etc. Is it fine? n minus 1 factorial is n minus 1, n minus 2, n minus 3 factorial. This one will be having a factorial. I'll make your doubt clear only. See, when you say 5 factorial, what is the meaning of 5 factorial? Say 2. 5, 4, 3, 2, 1. Can I write this like this? 5 into 2. This whole thing is a 4 factorial. Can I not also write it like this? 5 into 4 and this whole thing is 3 factorial. Can I not write this as 5 into 4 into 3 into 2 factorial also? That is what I use here. So I broke the higher number of factorial in terms of a lower number of factorial. Why? Because n minus 3 factorial was sitting in the denominator, which had to cancel it out. Okay. Now, what about if the difference of the difference came out to be a geometric progression? Okay. Something in this case. So here is a problem where they have asked you for sum of, sorry, nth term also and sum of n terms also. So how do I solve this kind of a question? So the problem is nth term, how do we find that out? So here you realize that if you take the difference, you get 2, 4, 8, 16. Now, this is a GP. Now, so here is a theory once again, which I would like to share. Theory is if your given terms themselves are in GP, then your nth term is given by AR to the power of n minus 1. We all know this. Okay. We all know the nth term of a GP is AR to the power n minus 1, where R is the common ratio in all. Right. But if your delta 0 is not in GP, but your delta 1 is in GP. Okay. Delta 1 is in GP. Then in that case, your nth term becomes no down AR to the power n minus 1 plus a linear expression in n. Again, I'm giving you this without a proof. Proof is not important here. Okay. Where this R is the same common ratio which is there in this GP. Whatever is the common ratio, that R will be present over here. So it will be a mix of an exponential term over here along with a linear term. So A, there will be some common ratio here, whatever it is it 2, 3, 4, 5, whatever is the common ratio to the power n minus 1 plus Bn plus A. Don't worry. ABC can be easily found out from the given terms themselves. Okay. If your delta 2 is in GP, if your delta 2 terms are in GP, then no down your nth term would be AR to the power n minus 1 plus a quadratic in n. Okay. Like that. So only thing that will change is this linear quadratic, cubic, bi-quadratic, pentic. Okay. Now, using this concept, we are going to solve this problem. First note this down. Okay. Done. Okay. Let's come back to this question. So in this question, you realize that delta 0t was not in GP, but delta 1t was in GP. So when delta 1 is in GP, what was the nth term I told you? Let's check. Let's check. Let's check. And the term was AR to the power n minus 1 plus Bn plus C. So it will become A. What is R here? What is the common ratio of this GP? Write it down on the chat box. 2. Okay. Plus Bn plus C. Okay. Now, all I need to do is know my A, B and C and my job for finding the nth term is done. So let's put n as a 1. When you put n as a 1, you get the first equation. When you put n as a 2, that is your second term, 3 is going to be 2A plus 2B plus C. When you put n as 3, that is your 7 will be equal to 4A plus 3B plus C. Correct? Three equations, three unknowns is all I need to solve this question. Okay. Anything that I missed out, do let me know. Okay. Take the difference of these two. So I'll give you A plus B is equal to 2. Take the difference of these two. You get 2A plus B is equal to 4. Again, take a difference. A will be equal to 2 itself. Then B will be 0 and C will be minus 1. Correct me if I'm wrong. So once I've got this, I can write down my nth term as 2 into 2 to the power n minus 1, 0n minus 1. That is nothing but 2 to the power n minus 1. Check. Am I right? If I put n as 1, do I get 1? Yes. If I put n as 2, do I get 3? Yes. If I put n as 3, do I get 7? Yes. Okay. So basically, this is right. Is this fine? Now, once you've got the nth term, you can use your sigma method to find the sum. So sigma trr from 1 to n is going to be your sum. So this is sigma 2 to the power r minus 1. Okay. By the way, minus 1, you can write separately. Okay. From r equal to 1 till n. Now, some of you would be wondering, sir, 2 to the power r special series you never give us. Is it like that? I never told you 2 to the power r series. Think carefully and then answer. Have I not told you? Have I not told you how to find the sum of summation 2 to the power r, r from 1 to n? Yes. No. Maybe. If I've told you, say yes. If I have not told you, say no. So what is your, what is your response person? Yes, I have told you. This is actually a GP. So I don't have to tell you a special series for that. So this is as we were saying 2 to the power 1, 2 to the power 2, 2 to the power 3, all the way till 2 to the power n. This is only told you with a GP, it's a GP, isn't it? So there's no need of a special formula for this. Why will I tell a special formula for this? Okay. It's a GP. Okay. So in a GP, the first term is 2. Common ratio is also 2. Number of terms is 1. Okay. So your result will be 2 into 2 to the power n minus 1 minus n. In short, you can write it like this, 2 to the power n plus 1 minus n minus 2. Okay. This is the sum till n terms. Again, never become complacent. Check. Check. Is it actually working or not? So if I put n as 3, I should get 11. If I put n as 3, I should get 11. Check. So 3 will give you 4 minus 3 minus 2. So 16 minus 5. 11. Yes, verified. Done. Okay. So this is undisclicked. Is this fine? How this method helps us to find the nth term? Once you know the nth term, the game is very simple after that. It's just sigma method of summation. Any questions? Okay. Now, most of you would be expecting that there is one more shortcut for this. Sorry. Newton never gave any method for such cases where your difference comes out to be GP. His method is only applicable when your difference or difference of difference or difference of difference of difference, whatever is the case. Okay. That comes out to be an AP or eventually becomes a zero. That can only happen when they're eventually becoming real. So he never gave a method for this. So this is the only shortcut known for this kind of a problem. Is it fine? Any questions? Okay. So with this, we are now going to move towards another type of series, which we normally call as AGS. AGS. A stands for Arthmetico. G stands for geometric. C stands for series. AGS. What we call is Arthmetico geometric series. Okay. Why this call Arthmetico geometric series is because in such kind of an expression, you would realize that there will be a term which will be growing like an AP. And another term of the same expression will be growing like a GP. Okay. See here. I've given you a series where such kind of observation is seen. Okay. So if you see the first, let me write a one for the sake of understanding it. Just see these A, A plus D, A plus two D, A plus three D. How are they growing? They're growing like an AP, isn't it? But at the same time, C1, R, R square, R cube, these are going like a GP. So this is a series where you realize that a part of the term is behaving as if they are growing in AP and a part of the term is behaving as if they are changing as a GP. Okay. Let me give you a live example. I have, I think one question based on the same. Okay. Let's say this question. Okay. Don't worry. I will solve it. But first, look at the series given to you. One, I'll write it down for you. One. In fact, I'll write it in a very subtle way so that you can identify what is happening. Okay. Same terms I'm writing. Don't worry. But I'm writing in such a way that you can identify what is happening. Now see, look at this one, look at this four, look at this seven, look at this 10. How are they growing? How are they related? What kind of a progression they are following? AP. Okay. And look at this term. One, one by five, one by five square, one by five cube, etc. How are they going? GP. That is why this type of series is given a special name, AGS, Arthmetico geometric series. Okay. Now, fine. You have understood what is this AGS. How do I sum this up? Okay. So here is a process which I'm going to discuss with you. And this process is a universally applicable process, which you can use to solve all questions related to AGS. So this process is very similar to what you have done in finding the sum of a geometric progression. Okay. If you recall, what was the process of finding the sum of N terms of a geometric progression? The same process is going to be applicable here as well. So let me just run you through that process. Okay. Ready? Let's do this. So how do I find the sum till N terms? As you can see, they're asking us for some till N terms and some till infinity. Let's do the sum till N term first. Then I will come to some till infinity. So first of all, let's write down correctly the Nth term. That means the last term till wherever we are summing up. Who will write down the, who will tell me the last term, the Nth term? Please write it on your chat box. Tell me the Nth term of this particular sequence. Be careful while writing it. Many people make mistakes. Nth term, I want. They are very smart. They have not mentioned the Nth term because then they know that some people will make mistakes there. Brilliant, Situ. So Situ says 3N minus 2 by 5 to the power N minus 1. Agreed everybody? Don't trust him. Just put N value as 3. Are you getting 7 by 5 square? You will be getting. Put N value as 4. Are you getting 10 by 5 cube? Yes, you will be getting. Okay. So absolutely right. Anybody who has a doubt how he got this? You know how to find Nth term of an AP, right? You know the GP sequence also, right? So you already know. So now the process that we had learned while finding the sum of a GP was we used to multiply with the common ratio of the GP, right? I am going to do the same thing over here. I am going to multiply this SN with the common ratio of the GP involved. Okay. So what are the GP involved? 1, 1 by 5, 1 by 5 square, 1 by 5 cube. That is the GP part of your terms. So the common ratio there is 1 by 5. So I will multiply the whole thing with 1 by 5. Okay. Please note that I don't have any business with the AP part of it. AP common difference is of no use. The GP's common ratio is of use to me. Okay. So if you multiply the first term with 1 by 5, it will become 1 by 5. Write it one shifted to the right. How do you find the Nth term? Okay. Okay. I'll come back to that. First, let me complete writing it. See, Pratej, if I ask you this question, 1, 4, 7, 10, etc., if I write this sequence like this, what will be your Nth term of the sequence? What will your answer be? Just answer this question. 1, 4, 7, 10, what will be the Nth term of the sequence? Do it on your notebook and tell me. It's an AP, right? What is the Nth term of this AP? First term, N minus 1 into common difference. No. Nth term of an AP, we all know it, right? So on simplification doesn't become 3N minus 2. So that is what will happen to these terms. 1, 4, 7, 10, it will become 3N minus 2. And if you see the powers of 5, they're growing like this. 5 to the power 0, 5 to the power 1, 5 to the power 2. So whichever position you are at, the power of 5 is 1 less than that position. So if you are at Nth position, the power of 5 will be N minus 1. Simple. Now, recall in a GP, we used to subtract these two results, isn't it? So if you subtract these two, you'll end up getting 4 by 5 Sn. This will be 1. This will be 3 by 5. This will be 3 by 5 square. This will be 3 by 5 cube and so on. This will also be 3 by 5 to the power N minus 1. And the last term will be negative 3N minus 2 by 5 to the power N. So what did I achieve by doing this? Now, notice here, if you look at these terms over here, they are actually coming in a GP. Yes or no? Yes or no? Right? So what is the first term of that GP? 3 by 5. What is the common ratio of that GP? 1 by 5. How many terms would be there? Can anybody tell me? How many terms are there from this position to this position? Can somebody tell me? How many terms? Okay, options. Is it N terms? Is it N minus 1 terms? Is it N minus 2 terms? N minus 1 terms, correct. Okay, now see, many people wrongly take 1 also be to be the part of the GP. Note that 1 is not a part of the GP. Right? Don't make that mistake. But sometimes it may happen that the first term can also be taken into consideration as a part of that GP. But in this case, it is not. Your GP only starts from second term here, 3 by 5 onwards. Why? Because if 1 was a part of a GP, then the common ratio from the first two terms would have been 3 by 5. And from the second and the third term that would have become 1 by 5. Third and the fourth also 1 by 5. So first term is an odd man out. Don't include it. Okay? So here, 4 by 5 Sn will become 1 plus. Now, some of a GP is already known. So A, 1 minus R to the power of total number of terms divided by 1 minus R. And of course, the last term is 3N minus 2 upon 5 to the power N. Okay? Let me simplify this. In fact, I will copy this result somewhere over here. This is not required. This is just to explain Prateh Chandra in it. So just copy that same expression over here. So 4 by 5 Sn is equal to 1 plus. I think I had written this. So for the simplification, this becomes 4 by 5 Sn. 1 plus. This will become 4 by 5 and there's a 3 by 5 on top. So that will become 3 by 4. 1 minus 1 by 5 to the power N minus 1. Okay? Now, just make S the subject of the formula. Make S in the subject of the formula by multiplying with 5 by 4. 5 by 4 will give you 3N minus 2 upon 4 into 5 to the power N minus 1. Correct me if I'm wrong. So this becomes the answer to this series. Okay? Any questions? Any concerns here? So just remember this process that whatever you had used for finding the sum of N terms of a GP, the very same concept has been used here to find the sum of an AGS also. Good enough? Any question? Any concerns? Okay. Now, how do I find the sum till infinite terms? Okay, how do I find the sum till infinity? So for infinity, please note down the following things. You have to take a limiting case as N tends to infinity of this Sn. Okay? In short, you have to take a limit as N tends to infinity of this expression 5 by 4 times 15 by 16, 1 minus 5 to the power N minus 1, minus 3N minus 2 upon 4, 5 to the power N minus 1. Okay? Now, see how do I take the limit? Very important limit concept also. See, when N is a very large number, what will happen to this term? Can somebody tell me 1 by 5 raised to a power, very heavy number? What will happen? What will let term tend to 0? Exactly, Shalini. This will be 0. Okay? Now, what will happen to this term? Can somebody tell me? You must be thinking, are you saying that is infinity by infinity? No? So we have to evaluate it. Correct? But remember one small thing of mine, very simple concept I am going to give you and it is very obvious also. There is a polynomial on the top and there is an exponential in the denominator. Exponential functions are much more rapidly growing functions as compared to polynomials. Okay? So this infinity is infinitely more than this infinity. So this result will also become a 0. Always remember a polynomial by an exponential function. Both are having, you know, infinity, infinity. Exponential functions will be much more larger in value as compared to the polynomial. So as a result, everything other than 5 by 4 and 15 by 16, they will all go a 0 and this expression will lead to, if I am not mistaken, 20 by 16, 35 by 16. So this will give you 35 by 16 as your answer. Okay? Now, if you don't like this limit approach, I'll give you one more approach to do it. Don't worry. Okay? But this is a very interesting observation. In fact, you should note it down somewhere. It's an important part of the limit concept that a polynomial in n or a polynomial in x divided by an exponential in x, if your x is tending to infinity, then this result will be a 0 because exponential functions grow much rapidly. That means they take much higher value as compared to a polynomial we'll take. Getting the point? So polynomials are like super fast trains. Sorry, yeah. Exponentials are like super fast trains. Polynomial is our normal, you can say, good strain going slowly. Okay? Is it fine? Okay. So as I promised you, I'll tell you one more way to do this sum till infinity. So till sum till infinity, you start with the same approach. Okay? So I'll give you another way to get the sum till infinity. So alternative way to get the sum till infinity. In this approach, let me first write down my terms. What were the terms? I think this were the terms. Okay? Don't write anything. No need to write any last term. Just put dot, dot, dot because it's going till infinity. Right? Same approach. Multiply with the common ratio involved and write it one shifted. Okay? Dot, dot, dot. Take the difference. This will be 1. This will be 3 by 5. This will be 3 by 5 square. This will be 3 by 5 cube. Dot, dot, dot. Okay? Now from here onwards, you will end up getting an infinite GP. So infinite GP, we know the sum is A by 1 minus 1, 1 minus R. R is 1 by 5. So this gives you, I'm just simplifying it. This gives you 1 plus 3 by 4. Isn't it? That's nothing but 7 by 4. Correct? Let me know if I missed out anything. Okay? So S infinity will be 35 by 16, which is the same answer as what you brought over here. 35 by 16. Right? So there's no limit and all you have to take. You can directly use your infinite GP formula to get it. Is this fine? I have purposely missed out a step over here, but if you want, I can write it down again. So 4 by 5th S infinity is 7 by 4. So S infinity is 35 by 16. Is it clear? Any questions? Any concerns? Do let me know. Now the last part of our discussion, we'll be talking about those cases where sigma method of summation is still not going to work. Right? One of the case was what we saw a little while ago. And there's one more case where sigma method of summation is not going to work. Okay? And that case is what we call as series where you can use VN method. Okay? So there are certain series where the sigma method of summation, whatever we have learned, that is not going to work. In fact, AGS, it is not an AGS also. Okay? It is not any special series also. It is not an AP also. It is not a GP also. Okay? So I will take first few examples. Okay? From there, we'll get an understanding what method I'm talking about, what type of series I'm talking about, and what is this VN method that is going to help us in solving that. Okay? Let's take an example here. Let's say I want to find out the sum of 1 upon 1 into 2, 1 upon 2 into 3, 1 upon 3 into 4, 1 upon 4 into 5, and so on up till n terms. Okay? How will you solve this question? It's not an AP. The difference also is like not an AP or a GP. Right? And moreover, we cannot use our sigma method also. See, please understand here, our term here is 1 by r r plus 1. Right? If you're thinking I can solve this question by using this formula, let me tell you it is ridiculously incorrect to solve it. Ridiculously, yes. People will laugh if you do this problem like this. Okay? So please note this is not going to work. Are you getting my point? This is not going to work. You can't say sum of fractions is sum of the numerators together and sum of the denominators together. Does it happen like that? No, right? Right? So the problem here is of a different nature altogether, where sigma method of summation has failed miserably. It is not an AGS. It is not any special series. Then how else are we going to solve this? Okay? Now, method of difference is going to help us, but not the same way as what we discussed a little while ago. Okay? So there's a small change in that approach. So let us discuss this. Now, I will not erase this. I will keep it on your notes itself because I want you to keep looking at it whenever you are referring to your that never try doing this for this kind of a problem. Okay? So what do I do for this kind of a problem? The approach is you write down t r that is your rth term as difference of two expressions which have come from the same function. But one is having an r in the variable and other is having an r plus 1 in the variable. Okay? In short, can I not write this term as like this? Pijas Vinayam, what are you asking? Why this is not going to work, which I've crossed out? Are you asking that? Okay, I'll ask you a question. 1 by 2, 2 by 3, 3 by 4, 4 by 5, etc. If I ask you to sum this up, will you ever do this? 1 plus 2 plus 3 plus 4 by 2 plus 3 plus 4 plus 5? Do you solve this question like this? Do you do this to solve it? Definitely not, right? Some of fractions doesn't mean some of the numerators separately, some of the denominators separately are divided. In the same way, if I'm summing up 1 by r r plus 1, you don't sum the ones and you don't sum the denominators separately, right? It doesn't work like that. It's not allowed in mathematics, isn't it? Okay, that is why we cannot use this one. All right, now let's focus on what we can use, what about what we cannot use? Okay, so here what we do, we break this up as a difference of two such terms, okay? Where one is in terms of r, see both coming from the same function 1 by x. So normally we call this as a function v, okay? So if you put your x as an r, you get this term. And when you put x as an r plus 1, you get this term. So both are coming from the same function vx. That is why we call this type of method as a vn method. Now it may sound very weird because people will say, says somebody kept its name as vx. So we started calling it as vn method. Yes, that's how it worked. Okay. So you know, had the person who started working with this had called it by some other name, we could have called this method by some other name. Okay, anyways. So if you see here, we have written the rth term as difference of two such terms where both the terms that I have written has come from the same function. If you want, I can write x in the bracket also not an issue because normally you have a habit of recognizing it as a f of x, isn't it? Okay. Now what is the benefit? If you write your rth term like this, what is the benefit? Why are we writing it like this? Now see the benefit. The benefit is very similar to what we saw when we were deriving our special series. So your first term will become one by one minus one by two. Okay, just put r as one. Second term will become one by two minus one by three. Third term will become one by three minus one by four. And if you continue doing this, you end up getting one by n minus one by n plus one. Okay. Now add it. Remember, we had done a similar approach for our special series. So when you add t1, t2, t3, till tn, you're getting the sn. That is your required sum. Here, thankfully these terms will start cancelling off. Okay. Ultimately, everything will get cancelled off leaving our one and leaving our one by n plus one behind. And that becomes your answer to this question. Is it fine? So it is still a method of difference approach. But now the approach is we start writing it as difference of two such terms, which both come from the same function, but one has an input of r, other has an input of r plus one or r minus one. That depends on case to case. Okay. Now this part itself is the most challenging part of the question. So how do I figure out such a function? And how do I break it up like this? Many people say, is this the partial faction that you have used? In this case, yes, it is the partial faction. But only for this case, it is a partial faction. Maybe if I scale up the complexity of the if I make the problem slightly more complex, you may not call that as a partial faction. Okay. So this step itself, where you are breaking your rth term as difference of two such terms where one has been obtained by putting your variable as an r and other as an r plus one, that itself is the challenging part of this question. And what is the benefit of that? It basically leads to cancellation of many terms. Are you getting my point? Okay. So this is your answer to this particular question, which I asked you. And again, let's not get complacent. Let's verify our answer. If I put NS two, I should get half plus one by six. Half plus one by six, if I'm not mistaken, is two-third. So yes, if I put SS two, I get two by two plus one, which is two-third. So that is verified. Any question related to this particular solution? Setu, I hope your question is also addressed. See, what do we do? Why I'm calling it as a VX? Because both these expressions have come from the same machine. This machine may, if you put r, you get the first term or second term depends upon the case. And if you put r plus one, you get the second term. Why are we trying to write it as difference of two such terms, which are coming from the same machine is because then only such cancellations will take place. This cancellation that you see, no, it is happening because of that nature. Okay. Don't worry. We'll take more questions from there. You will end up getting an idea of what basically this means. Okay. So as of now, think you have to write your rth term as difference of two such terms, where in one term, if you put your r as r plus one, you get the next one or vice versa. In this case, it is. It depends on situation or situation. In this case, I have written it like this. Please note that VX is always not one max. It is depending on situation to situation. See, examples will clear most of your doubts. Okay. But before I move to the next question, see, this is not a predictable way of solving the question. Most of you are trying to get a standard operating procedure. Sir, one shoe fit solve case. You can't use the same technique to solve all the question. In this case, it is working like this. Some other case, it will work similar to this, but with a different approach. We will take questions. Sometimes you learn by looking at solving more questions. It's like learning how to drive. The person who has come to teach you how to driving, what does he do? It just tells you this is paddle. This is your accelerator. This is your steering and goes off. No, he goes in. It takes you to the road. There you understand, oh, this bike is coming like this. I have to stop my vehicle. Or there's a turn. I have to slow down. That is how you learn things in life. When you see problems, no, you will start learning, oh, this is what a killer was trying to say. Are you getting my word? So wait, let me, let me take few more examples, then you'll understand. Now, see, many times the question setter may ask you what is the sum till infinite terms also? If I in this case, they may also ask you what is the sum till infinite terms? Now, all you need to do is to find the sum till infinite terms, take a limiting case as n tends to infinity of this answer. We all know how to evaluate this case, divide by n to the power 1 throughout, take 1 by n as 0, because as n tends to infinity, 1 by n will be 0 and the answer becomes 1. So, why I'm showing you this part as well, because they may ask you a question like this. Now, let's take another question just to make things clear. But before I move on, if you want to copy down anything from here or take down any notes, please do so and wait for few more examples, then you will understand what I mean to say. This vx part, I'll make it clear. Is it fine? I hope you have copied out, copied down everything that you wanted to. Okay, let me take another question. Find the sum till n terms of this, 1, 3, 5, 3, 5, 7, 5, 7, 9, up till n terms. Now, see, let's go step by step. Let us write down the rth term over here and I would request everybody to do, please write down the rth term of this particular series on your chat box. I'm sure some errors would happen. That's why I'm asking you to write it down, because many people do mistakes in writing down the rth term. So much time to write down the rth term. Come on, write, say 2. So 1 by 2r minus 1, 2r plus 1, 2r plus 3. Do you all agree with say 2? Is this the rth term? Write a yes if you agree with him. If no, then give your version of the answer. Okay, Shalini agrees. Yes, okay, everybody agrees. Great. Now, see, sigma method of summation is not going to work. So what is going to work? I have to write this rth term. Now, this is the problem statement, say 2, just to answer your question. Only problem statement I need to solve is I need to solve, I need to break this down rth term as difference of two such terms where in one term, if you change your r with r plus 1, you get the other term. As I said, think like that. Whatever you write over here in terms of r, that r you change to r plus 1, you get the next term, which is your double question mark term. Think like that. Okay. Now, this itself is a challenge, but I will help you out. See, just write the first two factors over here. Okay. And here you write down the last two factors. Now, in the factors themselves, you see that here if you change your r with r plus 1, you get this guy. And here if you change your r with r plus 1, you get the second guy. Do you all agree with me or not? So if you replace your r with r plus 1, you get the denominators. Correct. And now see, if you had taken the LCM, you would have got the same denominator as you have on the left-hand side. But on the numerator, you will get a 4. Check it out. So what I'm going to do, I'm going to put a 1, 4, 1, 4th over here. Are you getting my point? So what have I done over here? I have written this term very tactfully and very strategically as difference of two such terms. Just do the simplification. 1 by 4 will cancel out the four term that you will get on the top. Take your time. See, I've pulled off my hands. Use, simplify the right-hand side terms and see whether you're getting left-hand side or not. If no, then you tell me sir, you're wrong. See, if you would have taken the LCM over here, you would have got r minus 1, r plus 1, r plus 3, correct? So if I had not put a 1 by 4, let's say, if I just put a 1, then what will happen? I'll get 2r plus 3 minus 2r minus 1. That will eventually give me a 4 only. That 4 in order to make it 1, because I have a 1 over here, that is why I put a 1-4th on top of both of them. Correct? Now if the question is, sir, how did you think of this? Then I have only one answer to that. You require a bit of exposure. It's a reverse engineering that you're doing. It's very difficult to assimilate things. So when you simplify, it's very easy. But from the simplified part, you are getting the two components that is difficult. So this is something that I was talking about a little while ago, that if there was a machine where you had put an r to get the first term, in the same machine, if you put r plus 1, you'll get the second term. And that machine is called that Vx machine. You don't have to write that down. Don't worry. While you're solving that question, you don't have to mention what machine you're using. All the questions that are required is to break this function, break this rth term as difference of two such terms, where one term is obtained by putting, let's say r, then the other one is obtained by in the same function you're putting r plus 1. Correct? Okay. If you want to know what is VR over here, you can call this as a VR or Vx if you want to call it 1-4th, 2x minus 1, 2x plus 1. So if you put Xs r over here, if you put Xs r over here, you get this term. And in this, if you put Xs r plus 1, you get this term. Check it out. Okay. So both are coming from the same function, but your input is r in 1 and r plus 1 in the other. Now, what is the benefit of this? Benefit of this, as I already told you, it leads to cancellation of terms. That is the benefit. Okay. Let's complete this question. Setu, is this clear? Should I raise this? Do you need to note it down? So I'm raising it, right? Now, you don't have to show this Vx to the examiner. Don't worry. It is for your own internal understanding that I told you like this. Okay. So once your TR term is written like this, you start putting r as 1. So when you put r as 1, you get 1-4th by 1 into 3 minus 1-4th by 3 into 4, sorry, 3 into 5. Okay. Second term will become 1-4th, 3 into 5 minus 1-4th, 5 into 7. Third term will become 1-4th, 5 into 7 minus 1-4th, 7 into 9. Okay. And so on and so forth. So your last term will become, sorry, I may extend the class by a couple of more minutes. Don't worry about that. This will become 2n minus 1, 2n plus 1 minus 1-4, 2n plus 1, 2n plus 3. Okay. And once you add this, on the left-hand side, you get the required sum. And on the right-hand side, there would be a cancellation of many terms. This cancels with this. This cancels with this. Similarly, this will also cancel with the term coming next to it. And eventually, this will also get cancelled with some term prior to it, which are not written. Eventually leaving you with 1-4th into 1 by 3, which is 1 by 12. And this term, which is 1 by 4, 2n plus 1, 2n plus 3. This becomes your answer to this question. Is this fine? This requires practice. Okay. This is the last part of our chapter. This requires a bit of practice, VN method. You can find some problems in RD Sharma also, towards the last part of an exercise. Then you can start working on your modules and the DPPs. Now, here also, if I ask you what is S infinity, can you find it out? Can somebody tell me what will be S infinity in this case? Write the answer on the chat box. 1 by 12, right? Say it the way. Absolutely correct. Correct. So, if you have to take the limiting, sorry, if you have to take your limit of this as n tends to infinity, please note that this will be tending to 0 because it is denominator heavy. Denominator is all infinite terms. So, leaving you with 1 by 12. Is this fine? So, with this, we close this chapter. Next class, we will be starting with PNC, permutation and combination. So, please practice at least this VN method. That is very important method for us. Okay. Over and out from my side. Bye-bye. Take care. Good night.