 All right. So let's get started. So as I already told you the topic for today's discussion would be progressions and series, progression and series. Okay. I think I already told you the difference between progression and series. By the way, I'll just try to recall this for all of you sequences, progression and series. What are the difference? So what are the difference between a sequence, progression and series? Okay. Just to quickly write it down for you all. I will do that in class, no worries because it requires a bit of explanation and doesn't take much of a time. So in order to maintain the continuity, I'm doing this right now. I'll definitely do that in class once we meet face to face. Okay. Okay. So what is the sequence? Basically it's the succession of numbers. So let me write it down as a definition of succession of numbers. Arrange in a definite order. Please mute your mic everyone. According to some well defined law or rule. Whereas what is a progression? A progression is basically a sequence which can be described by. So basically if the terms of the sequence, if the terms of a sequence is described by an explicit formula. Okay. Then you can call that sequence as progression. So what is the progression? It's a special types of sequence only. So sequence you can say it's a bigger set sequence is a bigger set. Okay. And two sequences where you have an explicit formula to represent the value of the term will be called as the progression. Okay. Let me give you some example in order to explain this sequence when we like a sequence of prime numbers, a sequence of prime numbers. Now here we don't have any formula right now which tells you to find out the value of a prime number given its position is known. Right. But progression, like there is an explicit formula by which you can find out the value of the term given its position is known. Right. For example, let's say numbers like one, three, five, seven, etc. Basically this is a progression because you know that any generic term could be generated as two and minus one. So put n as one, you get the first term, put n as two, you get the second term like that. Okay. So this is the formula for the nth term in the sequence. So all progressions are sequences, but all sequences are not progression. Okay. By the way, Fibonacci sequence, can you call Fibonacci sequence as a progression also? Anyone? All of you know Fibonacci sequence right? How does the Fibonacci sequence go? Zero, one, one, two. It goes like one, one, two, three, five, eight and so on. Do you know that there is actually a formula for finding the nth term of a Fibonacci sequence that is called the binnit formula? You can note it down. Binit formula. You can find out the nth term of a Fibonacci sequence by using this formula one by root five. One plus root five by two to the power of n minus one minus root five by two to the power of n. Okay. So this formula, you can find out the nth term of a Fibonacci sequence. So Fibonacci sequence can also be called as a Fibonacci progression. Okay. Now what about series? What is series? If you add the terms of a sequence, it becomes a series. Okay. So if you add the terms of a sequence, right, then you get a series. Okay. So primarily in this chapter, we are going to talk about progressions and series only. Okay. Sequences, basically we are not going to cover up because it's mostly a part of your NTSC and all syllabus. I think most of you would have written NTSC last year. The mental ability part, we have a lot of sequences questions, but we are not going to talk about that. We are only going to talk about those kinds of sequences where there is a presence of an explicit formula for it. Is that fine? So any doubt regarding what is the difference between sequence progression and series? Okay. Sir, could you go back a minute? Sure. This note will be shared with you. Don't worry about that. Was that Krishna, right? No, sir. Not me. Can I go to the next slide now? Yes, sir. So in our course of study, we are going to talk about primarily four types of progressions. Okay. So the progressions that we are going to study in our class 11th curriculum as well as in J-Main are the following. One is the arithmetic progression. I'm sure you would have been exposed to these kinds of progressions already in class 10th. Okay. Second is geometric progression, which probably you would have studied in class 11th. Harmonic progression, which I don't think so you would have covered in school because harmonic progression is not a part of your school curriculum, neither in CBC nor in ISC. And the fourth type that we are going to touch upon is called the arithmetic geometric progression. Arthmetical geometric progression. Okay. So let me begin with arithmetic progression, which you probably are aware of. It's slightly faster while talking about arithmetic progression. So what is an arithmetic progression also called as AP? Any succession of terms where the subsequent term differs from the preceding by a common difference like this. So basically AA plus D, A plus 2D, etc. Where here A is your first term. A is your first term. Also you can call it as T1. Okay. D is the common difference. What we call it as common difference. Please note that common difference is always the subsequent minus the preceding. So it's always Tn plus 1 minus Tn. Are you getting my point? Don't do Tn minus Tn plus 1. And this common difference must be true for all n. It must be true for all n belonging to the series. Okay. It should not happen that the common difference is fixed for the first five terms and then it is changing for the next six terms. So it should be fixed throughout all the positions of the terms. Okay. So if you continue doing this. Okay. We get arithmetic progression. So this is your T1. This is your T2. This is your T3. This is your T4. Can somebody tell me what will be your Tn? Where n is any general position? A plus n minus 1 into D. Right. A plus n minus 1 D. You have been doing this since last year. Right. Nothing new to you. Correct. So this is what we call as the nth term of an arithmetic progression. Okay. Nothing new here guys. This is something which you are already aware of. Okay. Now how do we find the sum of an arithmetic progression? Now. Long, long ago in a German school there was a six year old student who was given a punishment by the teacher to sum up natural numbers from 1 to 100. Okay. Now this German child, he was very smart. What he did? He wrote the same series. Now guys be watchful about the terminology has been used. Now I'm calling it as series because I've added it. Okay. So what he did? He wrote the same series in a reverse order. So he wrote 100, 99, 98, all the way till 1. Okay. And he added it in this way. He added S with S. So 2S. He added 1 with 100. That's 101. He added 2 with 99. That's again 101. He added 3 with 98 which is again 101. And he kept on doing it and he realized to his surprise that all the terms which were equidistant from the beginning and the end were adding up to a fixed value and that sum was actually the sum of the first and the last terms of this particular series. Correct. This made his life super, super easy because he could easily write this as 101 times 100. Okay. So S is nothing but 101 into 50 which is 5050. Okay. So this made his life very, very easy and he could find out the summation of this particular series. Now this child later on was popular by the name of Karl Fetrich-Goss. Karl Fetrich-Goss. I'm sure you would have heard of his name. If not, then you'll soon hear his name in the next year when you're doing electrostatics. I'll be teaching electrostatics. Yeah, Dheera sir will be teaching. Did you hear a sound next to me? Did you recognize the voice, the man himself? Dheera sir. Okay guys. So the same logic has now been applied to sum up any AP in general. So let's say I have an AP whose terms are A, A plus D, A plus 2D, all the way till let's say A plus N minus 2D till A plus N minus 1D. Okay. Just go away from here. Yeah. Yeah. Now the same logic has been now used here. We'll write the same series in a reverse fashion. We'll write the last term in the beginning. Okay. The second last term here. Third last term here. And so on till we reach A. Okay. Let's add this two series together. So SN and SN will become 2SN. This will again be 2A plus N minus 1D, which is nothing but the sum of the first and the last term. And you'll keep on getting 2A plus N minus 1D all the way till your last position. Okay. Okay. In short, you will have N times 2A plus N minus 1D. Okay. So your SN will be N by 2, 2A plus N minus 1D. I want to know from the ISC guys, Anurag, Skanda and Shankin. Have you done this chapter in school? Yes. Yes. Okay. Did you find this chapter easy? Moderate? Difficult? Easy only, right? Okay. That's why I was going slightly faster because I'm assuming that you've done this in school or you've done this last year. Is that fine? So please make a note of this because this is going to be the sum of your N terms of any arithmetic progression. Is that fine? Can I move on? In case of any doubt, any concern, please feel free to stop me. Okay. Now, let us do a bit of analysis about this formula. Okay. So we'll analyze this formula a little bit more before we move on to any kind of problems on it. A bit of analysis about the sum. So the formula for sum of N terms is N by 2, 2A plus N minus 1D. Right? You can write this formula also as N by 2, A plus A plus N minus 1D. Correct? Now do you realize that A is actually your first term and A plus N minus D is actually your nth term of the sequence? That you can also call it as the last term of the series. Right? So you can say that the sum of N terms could also be found out if you know how many terms are there and you know the first and the last term of whatever you are summing up. Yes or no? In a similar way, you can also extend this as the sum of the second term and the second last term. Isn't it? Because now you can also break this up as A plus D and A plus N minus 2D. Correct? And you can keep doing this. You can write it as N by 2, T3 plus TN minus 2 as well. Correct? So if I keep on going, can you complete this formula? If I have a TK, what will I have as the other term, please? Anyone? Type it out if you want to type it. Fast, fast, guys. If let's say I take any kth term, then what should I have the next term, T? What? N minus K plus 1. No. Yeah. Who was that? N minus K plus 1. Correct? Is that fine? Yes, sir. And if at all, if there is a middle term, it will be nothing but twice of the middle term. Okay. So it is N by 2 into twice of the middle term. Let me write it as a middle term. Is that clear, everyone? Any questions? No, sir. No, sir. Okay. Then, next analysis that I would like to present before you. So if you look at this formula again, N by 2, 2A plus N minus 1D. Okay. Let me break it up as 2A minus D and ND. Okay. Let me open the brackets. So it will become N square D by 2. Okay. Plus N into 2A minus D by 2. Right? Let me assume this to be A for the time being. And let me assume this to be B for the time being. You realize that the sum of N terms of any arithmetic progression will always have this kind of a structure. Right? So please make a note of it. The sum of N terms of an arithmetic progression will always have a structure of A N square plus B N. Right? Now, you must be thinking, what is so great about this analysis? Because you can always figure it out by expanding it. The great about this analysis is, in past, we have got a question like this. So let me pose that question to you. If the sum of N terms of any series is given as A N square plus B N plus C, is the series in arithmetic progression? Is the series in arithmetic progression? Guys, I'm asking this as a question to you. Take two minutes and try to answer yes or no. If no, why not? If yes, why? Take two minutes, guys. Don't rush. Sorry, I opened something which is blocking your view, I guess. So Hariman says, no, it looks like a quadratic equation. Didn't this look like a quadratic to you, Hariman? I think this also is a quadratic in N. So why there is a step treatment with this quadratic? So Anjali says yes, it's a series which signifies the sum of terms in an arithmetic progression. Aditya says no. Okay. Aditya, first, I would like to hear from you. Why don't you think that it's going to be an AP? Can you unmute yourself and just talk? Sir, it can't be an AP, but then the condition should be a C should be zero. Okay. Very smart. Okay. Let's say C is not zero. Then tell me. Of course, I've written C assuming the fact that C is not zero. I would have not written it at all. Okay. Aditya, let's say C is not zero. Then tell me. Nuraag has some different point of view. Okay. If C minus B is B minus A, wouldn't it become an AGP? No. A plus C is 2B. A plus C is 2B. Yes, only the C is... Okay, guys, I'm getting different versions of... I mean the same thing. Let me just clarify this to all of you. See, when you call a sequence as an arithmetic progression, when you know that TN minus TN minus one is a constant. Am I right? Yes or no? That is, it is independent of... It is independent of N, correct? And a constant, correct? Now, let us do an activity by which we can find out TN minus TN minus one. So let me first ask you, how would I get TN first? Given that, I know the sum till end terms. Who will tell me? How do I get a TN? If I know the sum till end terms, how do I get TN, guys? It's a question to everyone. Yes, anyone, please. Sir, SN minus SN minus one. Absolutely, absolutely. Who was that, by the way? Aditya sir. Aditya, very good. So Aditya says it will be SN minus SN minus one. Brilliant, absolutely correct. So according to my given formula, SN is AN square plus BN plus C, right? And SN would be what? AN minus one square plus BN minus one plus C. Correct? No. So I can write this as A times N square minus N minus one square plus B times N minus N minus one and C and all will go for a toss. So you will be gone from the expression, right? If you simplify this, it becomes A times 2N minus one plus a B. Any doubt so far? No doubt, correct. So if this is your TN, my dear friends, what will be TN minus one? Can I say TN minus one will be A times 2N minus three plus B? If you want to know how, why, when? I just replace my N with N minus one, that's it. Okay? Now if you do TN minus TN minus one, you will get A times 2N minus one minus 2N minus three and B will go off for a toss, which means you'll get a 2A. Okay? Now this gives you a false feeling that, oh, my TN minus TN minus one difference is coming out to be a constant which is independent of N and hence this series signifies the sum of terms which were in arithmetic progression. Now, wait a moment, there is a surprise coming up. Please note that this should be true for all N, right? Then only it will be an arithmetic progression. Now, when you do T3 minus T2, what do you get? What is T3 minus T2? You'll probably end up getting a 2A, right? Now try doing T2 minus T1, T2 minus T1. By the way, what is T2? T2 is nothing but sum of the second term minus first term, this is your T2 and T1 will be your first, sum of the first term itself, correct? Which is nothing but S2 minus 2S1, isn't it? So S2 is what, my dear friends? S2 is 4A plus 2B plus C. S1 is what? A plus B plus C, right? So let's do S2 minus 2S1, it will give you 4A plus 2B plus C minus 2A minus 2B minus 2C, correct? Now to your surprise, you'll see that you are left with 2A minus C as your answer. What does it indicate? Even though it is independent of N, the common difference does not remain same throughout. Here it is 2A and here it is 2A minus C. Yes or no? Are you making sense? So it makes me conclude that if a sum of a series is given to me as AN squared plus BN plus C, C being not 0 only the first term is not in arithmetic progression. The rest from second term onwards you will get the arithmetic progression. So whole series is not an AP. Are you getting my point? It is an AP only from the second term. Getting my point. So these kind of tricky questions can be asked to confuse you. Many people will conclude here. The why did not work for T1 because there is nothing called T0, right? So when I checked T2 minus T1 separately, I got a different answer. Are you getting my point everyone? So this will work only when N is greater than equal to 1, right? Sorry, N is greater than equal to 2. It is not working when N is equal to 1, okay? Are you getting my point? Yes, no, maybe type clear, do something so that I know you have understood it. Yes sir. Okay, okay. So don't get tricked by such questions. Fine. So let me do some problem solving. Enough of theory has been spoken so far. Let's do some problem solving. Okay, so let me start with this question. I'm sure you would have done this question umpteen number of times in class 10. Hope you can read the question. The question says the sum of N terms of two arithmetic progressions are in the ratio seven N plus one by four N plus 17. So basically the sum of two you can say SN and S dash N are given to you. Find the ratio of their nth terms. Find the ratio of their nth terms. So they're asking you TN by TN dash. Anybody who wants to tell their answer? I'm sure you would have done these kind of problems in class 10 itself. Isn't it? So what is SN by S dash N? It is going to be N by two 2A plus N minus 1D, right? Divided by N by two 2A dash N minus 1D dash, right? Let's remove the factors of N by two N by two and let's divide by two also. Let's divide this by two this by two. A plus N minus 1 by 2D by A plus A dash plus N minus 1 by 2D dash, correct? So this is given to us as seven N plus one by four N plus 17, correct? Now let us say since there is a clash of the name of the terms, let me call it as let's say I want to find mth term, the ratio of their mth terms, correct? Now if you see this expression carefully, okay? If you replace this guy with m minus one, right? Similarly here also you replace this with m minus one. What will you get? You will get A plus m minus 1D by A dash plus m minus 1D dash, correct? That means you are trying to replace n minus one by two with m minus one, which means you are trying to write n minus one as 2M minus two, which means you are trying to write n as 2M minus one. Yes or no? Correct? So what I am going to do is I am going to make the same replacements over here as well. So I am going to write this n as 2M minus one and this also as 2M minus one, right? So that this entire thing gets converted to A plus m minus 1D. Remember I am replacing this with m minus one. Denominator will convert to A dash plus m minus 1D dash. And on the right hand side of this expression, I will end up getting 7 times 2M minus one plus one by 4 times 2M minus one plus 17, which means I will get 14M minus six by 8M minus four plus this will be plus 13, right? Now all you need to do is replace your m with n back and this is going to be your final answer. Now don't tell me you have not done these kind of problems before. I am sure everybody has done this problem in class 10. It's the class 10 stuff. Isn't it? Is that clear? Anybody want to be in the explanation? Can I move on to the next question? Yes sir. So here goes the question. Find the sum of the first 24 terms of an AP. T1, T2, T3, T4 if it is known as T1 plus T5 plus T10 plus T15 plus T20 plus T24 is equal to 225. Please send me your answer privately. Please type your answer privately. Don't put it for everyone. No one so far? Absolutely correct. Rona, Krishna, Riva, Trippan, all of you are absolutely correct. Okay. Now let's discuss this guys. This is a very simple question. So you have terms from T1, T2, T3 all the way till T24. So we have to find this sum out, right? Okay. So I could find the sum out by using the formula number of terms 24 divided by 2 into sum of the first and the last term. Or sum of any two terms equidistant from the beginning and the end. Okay. So let's say I said T1 and T24. Okay. No Kirtanam. That's not right. Mayur, Ajay, you all are correct. Now here the best part is you have been given T1, T24, T5, T20, T10 and T15. Basically you have been given all the terms which are actually equidistant from the beginning and the end. Yes or no? So T1 is the first term. T24 is the last term. T5 is the fifth term. T20 is the fifth last term. T10 is the 10th term and T15 is the 10th last term. Isn't it? So all of them must have equal sum. Let's say I call the sum as X. So this will be also X. This will be also X. This will be also X. So indirectly from the given information, we can say that 3X is given to us as 225. No Shushant. Right, Anjali? So that means your X is going to be 75. So this is nothing but your T1 plus T24 itself. So all you need to do is plug in this value with 75. So your answer will be sum of N terms is 24 by 2 into 75. That's nothing but 12 into 75 which is nothing but 900. That's correct. So most of you have got this answer. Well done. I'm sure now you would have known your mistake. Is that fine everyone? Any explanation here? You may have solved it by other methods. I don't disagree with that. But I'm sure this is the easiest of all the methods to solve this question. Okay, so let's move on to the next question. Okay, so here you go. I'll read the question for all of you here. If N, the set of all natural numbers is partitioned into groups. Which is like S1 is 1, S2 is 2, 3, S3 is 4, 5, 6 and so on. Then find the sum of the numbers in S50. So basically it goes like this. S1 only has 1. S2 has 2, 3. S3 has 4, 5, 6. So if you go on all the way till S50, then what are the sum of the elements coming in S50? That's what the question is. Again, type your answer privately to me. No Vibhav, no Mayur, no Aditya, no Krishnan, that's not right. Anybody else? Take your time, don't have to rush through it. It is very clear that S50 will have 50 terms in it, right? But what is going to be the starting term? What is going to be absolutely correct, Kirtana? Kirtana, you are right. So at least one answer I have got, guys. No Aditya, it's much more than what you have said. No Ranganatha, I think you did some addition mistake. You are very close to the answer, by the way. No Sat Siddhartha, it's much more than that. One more minute I will give for people who are trying. Then I will start the discussion for it. Ajay from 12.25 to 12.75, let me tell you there are 51 terms. Okay, so let's discuss. Now you would have observed that the last term of a sequence is basically the sum of, let's say I talk about 3, 3 is the sum of 1 and 2, correct? 6, 6 is the sum of 1, 2 and 3. Are you getting it? Let's write the fourth term also, fourth term will be 7, 8, 9, 10. 10 is basically the sum of 1, 2, 3, 4, correct? So following the same pattern, can I say the last term would be, the last term over here would be the sum of terms from, sum of numbers from 1 to 50, right? Right, and what is that? That is 50 into 51 by 2, that is 12.75. No Tripan, I think you made a silly mistake. So this term here will be, the last term here will be 12.75, correct? If you go on 50 terms back, okay? This term here would be what? 1226, yes or no, right? So now if you want to sum this up, the sum of elements in S50 would be 50 by 2, 1226 plus 1275, correct? How much is that? So 1226 plus 175 will give you 2501 and this will give you 62525. This is going to be your answer. So I think Skanda is also correct. So two people only gave the right answer so far. No, Aranganath also gave later on. Well done guys. Kirtana, Aranganath, Skanda. Is that clear? Okay, let's move on to the next question. Yes, please do this. So basically the number of a positive integer having 3 digits are in AP and the sum is 15 and the number obtained by reversing the digits is 594 less than the original number, find the number. Anyone? Wait, sir, one second. Yeah, yeah. Take your time. Whoever has registered with the name of Mac, that is not the right answer. I think your digits are correct but you have not written it in a proper way. That's right. Netra, that's right. Now that's correct. Ravi Kiran also. Yeah, should I start the discussion? Sir, one second. Wait. Okay, okay. Correct, Aditya, correct Pratam. That's right, Kirtana. No, Aranganath. 933 is not in AP, right? First of all, correct Skanda. Correct. Okay, so most of you have solved this correctly. Well done. So guys, before I start solving this problem, a quick theory about selection of terms in an AP. Selection of terms in an AP. Okay, if you're asked to select three terms in an AP, we normally select it as A minus D, A and A plus D. I'm sure you would have done this in class 10th also. Correct? No Arpita. Yeshushan, that's correct. Correct, Aditya. Okay, now I'm solving this. All of you please pay attention. If you're asked to select four terms in an AP, we normally select it as A minus 3D, A minus D, A plus D and A plus 3D. Now what are the benefits of selecting terms like this? If you add them, you would realize that the D completely gets cancelled off. Okay? If you're asked to select five terms, it will be A minus 2D, A minus D, A, A plus D, A plus 2D. Correct? If you're asked to select six terms, it will be A minus 5D, A minus 3D, A minus D, A plus D, A plus 3D, A plus 5D, etc. Okay? Basically, the fifth term will follow from, the third term, sixth term will follow from the fourth term. So you just take a term to the left and to the right, you get the five terms of the series. Okay? Now, inspired by this small theory, I would like to solve this problem now. The digits are in AP. So let us take the digits themselves as, let's say the digits are, let's say the digits are, A minus D, A and A plus D. So these are the values of the hundred, tens and ones place digits of this three-digit number. Okay? Now, first information is that their sum is equal to 15. Correct? Which means, 3A is equal to 15, which means A is equal to 5. Okay? So middle term is 5 for sure. So this term is 5. Now, what is the number itself? The number itself is 100A minus D, plus 5 into 1050, plus A plus D, right? If you reverse this number, if you reverse this number, it will become 100A plus D, plus 50, plus A minus D. No Ronat, no Abhishek. It's Ulta, 852 is the number. Okay? Now, what is given to us in the question? The question says, the number obtained by reversing the digit is 594 less than the original number. So this plus 594 will be the original number. So I can say 100A minus D, plus 50, plus A plus D, is equal to 100A plus D, plus 50, plus A minus D, plus 594. Okay? 50, 50 you can drop it. Let us try to consolidate the terms over here. I think you can drop the factor of A also. Okay? 100A also will go off. So basically it becomes minus 100D, minus 100D, plus D, plus D is equal to 594. That means minus 198D is 594. So D is 594, or minus 594 by 198, which is actually negative 3. Correct? So once this is 5, this will be 5 minus of minus 3 and this will be 5 minus 3, correct? Which means the number is 852. So the answer here you must get is 852 is the number. Okay? Most of you have got this right? Well done. Any question? Okay. Now let us look into some properties of an arithmetic progression. Let's look into some properties of an arithmetic progression. Just everybody listen to this carefully. If A1, A2, A3, till An are in AP with a common difference with a common difference of D, okay? Then please note that even if you add or subtract the same terms from all the terms, let's say K, this will also be in AP. Can everybody tell me what would be the common difference? It will remain the same. Right. It will remain the same D. Many people get tricked over here. Okay. If you multiply every term with a K, let's say K is a non-zero constant and you are multiplying every term of this particular arithmetic progression with a K, will it be still in AP? Will this sequence still be in AP? Yes or no? Yes. It will still be in AP with common difference? 2. Right. Common difference is KD. KD. KD, right? Similarly, if you divide all the terms of an arithmetic progression by a non-zero constant K, please note that this will still be in AP with a common difference of D by K. D by K. Absolutely. Okay. Next property is if A1, A2, A3 till An are in AP. Okay. And B1, B2, B3 till Bn are also in another AP. So let's say there are two APs. Then please note that if you add or subtract, if you add or subtract the subsequent, if you add or subtract the corresponding terms here, this will still be in AP. But what will be the common difference? Let's say this has a common difference of D1 and this has a common difference of D2. Then what can I say about the common difference here? D1 plus minus D2. Are you getting it? But remember A1 by B1, A2 by B2, A3 by B3, etc. or A1, B1, A2, B2, A3, B3, etc. they will not be in AP. Getting this point. Another specialty about AP is if A1, A2, etc. till A and R in AP, please note that any Rth term could be written as the mean of R minus kth term and R plus kth term. So term equidistant from that term will always add up divided by 2 to give you the same term. For example, let's say I have 1, 3, 5, 7, 9, 11 like this. If I take, let's say this term 7, take any term equidistant from it. Let's say I go one this side and one this side. So 7 will be nothing but 5 plus 9 by 2. If I go to this side, so it will be 3 plus 11 by 2. Are you getting my point? So any term is nothing but the mean of the two terms which are equidistant from it on either side of it. Are you getting it? That's why we say that if A, P, C are in AP, B is nothing but A plus C by 2 which means A plus C is equal to 2B. Please remember this property. Many times in order to confuse you, they write the same thing as A minus B by B minus C is equal to A by A. Who is singing there? Shristi, you need to mute your mic. Thank you. Sometimes when I'm taking class at my home, my daughter will come screaming. So you can't help it out. So why I'm telling you this because there is a property which is of a similar nature coming in GP and HP as well. So please make a note of it. This has come in one of the exams. Is that okay everyone? Are these properties are clear in your mind? Okay. So can you take a few more questions here? I'm sure you can read this. Some of first 100 terms common to the two APs. So there are two APs here. One goes like 12, 15, 18. Other goes like 17, 21, 25. So you have to find the sum of the first 100 terms which are common to these two APs. Take your time. Please reply privately to this. Yes, Netra. You can always take it as AA plus DA plus 2D also. It doesn't matter. It's just that taking A minus DA and A plus D makes it convenient. Sorry, I was answering some other question of Netra not related to this question. Okay, sir. Thank you. Akshay, look at the options. Option is only given to you. It has to be one of the following. That's correct Vibhav. Correct Aditya. Correct Shankin. Correct Patam. Awesome, guys. You guys are awesome. Good. That's correct Anurag. Correct Aleman. Correct Ananya. Okay. That's correct Krishna. Let's discuss this. See here. First of all, let us figure out what is the first common term in both the APs. So if I continue writing the first AP. I can see I am going to write 21 next which is going to be your 21 present here also. Right. So basically 21 is the first common term to both the APs. Okay. Now can anybody guess when will be the next common term coming up? The next common term would come up at a common difference which is the LCM of the common difference of these two APs. Correct. So the first AP is in a common difference of 3. The second AP is in a common difference of 4. So the LCM of 3 and 4 which is nothing but 12. So 12 would be the next time you will get a common term. So 21, the next term will be 33. The next term will be 45. The next term will be 57 and so on. Is that clear everyone? Does that make sense? Okay. So first term is 21 and the common difference is going to be the LCM of the common difference of these two which is 12. Okay. So they are asking you the sum of 100 such terms. They are asking you the sum of 100 such terms. So what would you do? Use the sum of an AP formula 100 by 2 2 into 21 plus 100 minus 1 times 12. Correct. So what does this give us? This will give you 61500. Is that fine? Okay. So basically what you can do the calculation very easily here it's not very tricky. 2 you can cancel off from here and here. So that's going to be 100 times 21 plus 600 minus 6. So that's 100 times 600 and this will be I think 15. So it is 61500 which is option number. See that is correct. Well done guys. Good. Okay. Now these questions seem to be easy. Now let us do a challenging question. Okay. Let's see who gets that question. ABCD are distinct integers. Now read the question very carefully. They are distinct integers forming an increasing AP. What is the meaning of increasing AP? Increasing AP means every subsequent term is higher than the preceding one. Such that D is equal to A squared plus B squared plus C squared find the value of A plus B plus C plus D. Let's see who gets this right. Can please write down the answer privately to me? No, no, no, no. Shashank here D is not the common difference. Please note that A and D doesn't mean it has to be first time in the common difference always. Here ABCD are some generic terms. Okay. Please do not treat D to be the common difference. ABCD are certain four terms in an arithmetic progression. Okay. So you can call common difference to be some other name. Probably you can call it as capital D or something. The answer for this is a value, Abhishek. You'll get some value. Value always means a number. Okay. Read the question. Find the value of A plus B plus C plus D. That doesn't mean you'll get the answer in terms of something. You'll get a number for it. Yeah, that's the challenge Aditya. Okay. Anybody who's taking a try about to finish it off. I can give that's correct. Wow. Well done. Well done. But you wrote it for everyone to see that. Correct Krishna. So guys, that's the right answer. Answer is two. Hope you did not guess it. Did you guess that? Okay. So let's let's solve this question. So what I'll do is let me call a as your first term. Okay. Okay. And be let's call it as let's say a plus some common difference D. Now, since your small D is already used up, I have to use a different name for the common difference. Here I'm calling it as capital D. So let's see B a plus 2d and let D be a plus 3d. Okay. Let's use the condition given to us in the question that D is equal to D is equal to a square B square and C square. Okay. If you expand this term, you get a plus 3d is equal to 3a square. And from here you will end up getting a 6 ad and you'll get a 5d square. Correct. Now, write this as this. Okay. So basically you have written. I'm sorry. I'm sorry. Sorry. This is 3a square and this is a. Okay. Now treat this as if it is a quadratic. Sir 3a square minus a. I'm so sorry. Minus a. Yeah. Quadratic in D. Okay. And let us use our quadratic equation formula to get our D. D will be minus B plus minus B square minus 4 a C. Correct. All divided by 10. Isn't it? Till this step, it is clear to everyone. So what I did, I'm going to repeat my steps. First of all, I assume my BCD in terms of A and capital D. I use this equation over here. I use this equation over here. To reach a quadratic in D here. And then I use my Shridhar Acharya formula or the quadratic equation formula to get this. Now, since we know that the common difference is real. Right. Can I say this term, which is under the under root sign, this must be positive. So this term must be greater than equal to 0. Right. Let us simplify this. So we have 36 a square minus 36 a plus 9 minus 60 a square plus 20 a greater than equal to 0. Which on simplification gives you minus 24 a square minus 16 a plus 9 greater than equal to 0. Just multiply it throughout with a negative one so that, you know, you end up making the positive leading coefficient as positive here. And sign will reverse. Okay. Now, this is basically an inequality. Okay. And we know how to solve quadratic inequality. Remember your chapter's inequalities. We had learned that if a quadratic equation is factorizable like this. And you try to solve for less than zero. Then X should lie between alpha and beta. Let's say alpha is the smaller. Correct. In the same way, I'll find out the alpha and beta over here as well. So let us find the alpha and beta out. So here X is your A. Okay. And I will try to find out my alpha beta from here again by using Shridhar Acharya formula. So it's going to be minus B. Plus minus B square minus 4 AC, which is 36 into 24 by 48. Okay. So your two roots will be this. Yes. What happened? Any question? Yeah. We'll discuss your method also. Okay. So let me complete this. Now this will give you, by the way, this is nothing but 256 plus 36 into 24. What is that? All of you please calculate that 256 plus 36 into 24. That gives you 1120. 1120 is nothing but 16 into 17. 1120 is 16 into 17. Okay. So you can take out a factor of four outside. Okay. So when you divide by four, it becomes, sorry, divided, yeah, divided by 48 throughout. So it becomes minus one by three plus root 70 by 12. Correct? Plus minus. So you can say that your A lies between minus one third minus root 70 by 12 and minus one third plus root 70 by 12. Now A is an integer. Please read the question carefully. In the beginning itself, it was written that A, B, C, D are integers. Now tell me how many integers do you think will lie between these two numbers? Now remember root 70, you can take it as 8.5ish something or 8ish. Okay. So how much is this? So can I say zero will lie because one is negative and other is positive and minus one will also lie. Check it out. Are you convinced with that statement? Yes, sir. See this term is approximately minus 1.03. Okay. And this term here will be somewhere close to 0.36. So you'll end up getting a minus one in between and zero. So there are the two possibilities that your A could be either zero or negative one. Are you getting my point? Now let us take A as zero first. So there are two possibilities. One is A is zero and other is A is minus one. So let's say A is zero first. Okay. If A is zero, let us find out our common difference. Let's go back to this equation. And in this I'll put my A as zero. I'm going to put my A as zero in this. Tell me what is the common difference that you are getting? I will get a 3 plus minus under root. This will become a minus 3 square which is 9 by 10. So either D will be 3 minus 3 by 10 or 3 plus 3 by 10. Both are not possible. Why? Because your D has to be a natural number. It has to be an increasing AP. So D has to be positive. And from one integer to another integer if you are going, your D has to be some kind of a positive integer which is a natural number actually. So it means that A equal to zero itself is not possible. So let us now try A as negative one. If you take A as negative one, again let me put A as negative one in this. So in this again I am going to put A as negative one. Let's check what do we get? So this becomes a 9 plus minus 9 square. This becomes minus of 80 by 10. So it becomes 9 plus 1 by 10 or 9 minus 1 by 10. Again 9 minus 1 by 10 is not possible. But 9 plus 1 by 10 is possible which is actually 1. So what did you figure out that your common difference is also 1. So this is a possibility with a common difference of 1. So your B is going to be, so A is minus 1, B is going to be 0, C is going to be 1 and D is going to be 2. So your A plus B plus C plus D is going to be minus 1 plus 0 plus 1 plus 2 which is actually 2. So this was a slightly lengthy and challenging problem. Shashank I would like to hear from you. How did you solve this in your way? Did you get the same answer by the way first of all or not? Yes sir. How did you solve it? Sir you know that B minus A is equal to D minus C which is equal to the common difference. Okay. And when you rearrange the terms you get C plus B is equal to A plus D which is equal to the common difference. Sorry B minus A is equal to D minus C then. That is equal to the common difference. Okay let's say capital D then. And when you rearrange the terms you get C plus B which is equal to A plus D which is equal to common difference. One second. When you rearrange the term how by sending A on the other side? Yeah. Okay so what do you write? B is equal to A plus D then? No you write B plus C which is equal to A plus D. B plus C is equal to A plus D. Which is equal to the common difference. Oh no. These three are equal. But we cannot say this is equal to the common difference. You can just put transpose C to that side. See when you're transposing A you have to transpose everywhere right? You can't transpose only in two of them. Are you getting my point? Yes sir. Okay. Anyways. Well tried guys. Especially the ones who got it right good. Great. So next we are going to talk about, next we are going to talk about GP. Geometric progression. Yeah. So Shomik you can mute yourself please. Now what's the geometric progression? The geometric progression is basically a succession of numbers where the subsequent number is obtained by multiplying the preceding by a common ratio. So basically if you see AR is obtained by multiplying A with R. AR square is obtained by multiplying AR with R and so on. Right? So if this is your first term, this is your second term, this is your third term, this is your fourth term. Can somebody tell me what will be your nth term be? AR to the power n minus one. Is that clear? Okay. Now since common ratio is obtained by multiplying, taking the ratio of the subsequent to the preceding. Right? Please note that this implies that no term in a GP can be zero. This is very very important. No term in a GP can be zero guys. Because common ratio in that case will become undefined. Okay? So your common ratio R could either be from zero to one. It could also be greater than one. It could also be negative. It could also be complex or imaginary. That's fine. But it cannot be zero. So no term in a GP can ever be zero. Is that clear? Now how do we sum up a GP? So now let's talk about the sum of n terms of a GP. I'm sure you have done this chapter, sum of GP. This concept in school. Everyone? Yes, sir. Okay. So I'll be a little faster here. So let's say if you have a series whose terms are in a geometric progression. Let's say Rn minus two ARn minus one. Okay. Now how do we sum this up? The process here is that we multiply this entire series with R. R being the common ratio. So what I'm going to do is I'm going to multiply the entire series with R. So when you do that, A into R will become an AR. Correct? AR into R will become AR square. So I'm writing it one shifted to the right. Right? So if I keep doing it, this will become AR minus two. This will become AR minus one. And the last term multiplied with R will give you AR to the power n. Right? So of course I multiplied the entire series with R and I wrote it one shifted to the right. Now what I'm going to do is I'm going to subtract the two series. So when I subtract my left hand side, I end up getting Sn one minus R. On the right hand side, I'll get an A. Let's assume a zero to be present over here if nothing is there and zero to be present over here. So A minus zero is A. And these terms will start cancelling themselves out. Correct? And here I will end up getting a minus ARn. So Sn becomes A times one minus R to the power n by one minus R. Correct? You could also write it as AR to the power n minus one by R minus one. You could also write it as AR to the power n minus A by R minus one. Now you must be thinking I just opened the brackets. So what's so great about this? Basically if you see this term, it's actually last term into R. Correct? This was your last term. Correct? So you can also write it as LR minus A by R minus one. Is that clear? Yes. Okay? Now many books will suggest that you should use this formula when your mod R is less than one. That means your R is between minus one and one. And you should use this formula when your mod R is greater than one. That is when you're saying R is greater than one or R is less than minus one. Okay? Now my question here is to all of you, what will happen if R is equal to one? Don't you see this formula fails in that case because your numerator and denominator will both become zero in that case? Isn't it? Sir, all the terms of the GB will be the same. Yeah, but doesn't this formula fail in that case, Sankin? I know for sure that all the terms will be AAAA, right? Okay? So basically you should get NA, right? But when you're using this formula and you're putting R as one, isn't it becoming zero by zero? How is that happening? Can you justify the correctness of this formula in this case as well? How do you get any answer from this formula itself? Can anybody tell me? It's very simple, guys. Sir, we can use limits. Absolutely. Use limits. So when your R is one, your sum of n terms could be found out by taking the limit of R tending to one on this term. Or the other term. Doesn't matter. So it's basically eight times limit R tending to one, R to the power n minus one, which you can write it as one to the power n also. Doesn't this resemble? Try to recall. Try to recall your limits chapter where you had done a standard algebraic limit x to the power n minus a to the power n by x minus a, right? What was the result for that? The result for that was NA to the power n minus one, isn't it? So here in the same way, your R rule, your x rule has been played by R. A rule has been played by one. So the answer will be N1 to the power n minus one. And there's an A waiting outside, which actually gives you AN. So this formula is not wrong. This formula just becomes a special case when your R is tending to one. Are you getting my point? No, something very important. If any geometric progression has its common ratio modulus as less than one, we call that geometric progression as a convergent series. We call that geometric progression sum as a convergent series. What's a convergent series? A convergent series is a one where even if you sum up infinite number of terms in that, your result will be a finite one. One example that I would like to share with you is one half, one fourth, one eighth. If you sum this up till infinite terms, your answer will be just two, right? Because it's a convergent series. On the other hand, what's a divergent series? A divergent series is a series where if you sum the series to infinite terms, your answer will either be infinity or negative infinity. So when your mod R is less than one, you can actually find the sum of infinite terms for this. So you can find the sum of infinite terms for this particular geometric progression. And that will be nothing but limit n tending to infinity A1-R to the power n by 1-R. Now who will tell me what happens to this term when n is very, very large? So if your mod R is less than one and you raise it to a very, very high power, what will happen to this term? Zero. You will try it on your calculator. Just do half into half into half into half. Let's say one lakh times, 10 to the power 5 times. You start realizing that your answer in the calculator becomes so, so small that it is almost zero. So in that case, your answer will convert to a simpler expression which is A1-R, A by 1-R. Please remember this result also. The sum of infinite number of terms of a convergent geometric series is given by A by 1-R. Is that fine with everyone? Can I move on to the next slide? Say something. Okay. Now, before I move on to anything, let's do some selection of terms in a geometric progression. Just like we did selection of terms in an arithmetic progression, let's talk about selection of terms in a GP. So if you're asked to select three terms in a GP, we normally select it as A by R, A, A-R. If you ask to select four terms in a GP, we select it as A by R cube, A by R, A-R, A-R cube. Now guys, let me tell you, it is not a compulsion to do it like this. It basically just makes your life simpler. It's not like if you don't use this, you will not get your answer. Nothing like that. You can still get your answer, but it just makes your life easy. Okay. Especially where your information about the product of the terms is given to you. Similarly, if you're asked to select five terms, it is A by R square, A by R, A, A-R, A-R square. Six terms, A by R to the power five, A by R cube, A by R, A-R, A-R cube, A-R to the power five, etc. Okay. Now normally you won't get to select more than three or four terms. Okay. Let's take some questions on this. Okay. I'll start with a simpler question here. If sine theta, root two sine theta plus one and six sine theta plus six R. Let's say I change this question slightly R first to three terms of a GP. First three terms of a GP. Find the fifth term. Again, please respond privately. 324. See, look at the options. That's correct. Correct. Correct, Siddhartha. Come on, this is a very simple problem. You should not take more than no Ranga. Okay. Let's check. See, if there are in GP, we can say that the second term by first term is equal to third term by second term. The common ratio must be fixed. Correct. Look at the options. Look up from the options. So which means T2 square is equal to T1 into T3. Correct. This is actually a property, which is like if ABC are in GP, then the square of the middle term is equal to the product of the terms equidistant from it. I'll talk about it a little later on. Answer is 162 guys. Okay. Let me discuss it now with you all. Now T2 square means square of the middle term, which is nothing but root 2 sin theta plus 1 the whole square. This is equal to the product of sin theta and 6 sin theta plus 6, which means 2 sin theta plus 1 the whole square is equal to take out a 6 common. Okay. Now, let me cancel out a sin theta plus 1 term on both the sides. Now you must be wondering that I keep telling you that don't cancel out a factor from both the sides and now I'm canceling the factors out from both the sides. See, because I know that sin theta cannot be minus 1. Now you must be again wondering why can't sin theta be minus 1 because sin theta can be minus 1 also, right? Since sin theta becomes minus 1, it means this term would become a zero and even this term will become a zero, which means a GP has a zero term in it. Remember, I told you a GP cannot have a term in it as zero. That's not possible. Are you getting my point? So that's why I can safely cancel out sin theta plus 1 factor from both the sides. So I end up getting 2 sin theta plus 2 is equal to 6 sin theta, which means 4 sin theta is going to be 2. That means sin theta is going to be half, correct? So your first term is nothing but sin theta which is half. So I can write it like your T1 is half. T2 will be what? T2 will be root 2 into half plus 1, which is nothing but 3 root 2 by 2, correct? So from T1 and T2, I can easily figure out my common ratio and that will be nothing but T2 by T1, which is 3 root 2 by 2 divided by half, which is nothing but 3 root 2. So now your first term is known to you, your common ratio is known to you. Now let us try to answer what the question is demanding. The question is demanding from you the fifth term. Fifth term is AR to the power 4, which is half, 3 root 2 to the power of 4, which is half into 81 into 4. You can cancel out a factor of 2, so answer is 162, my dear friends. So option number C is the right option. Is that clear everyone? Is that clear? Can I move on to the next question please? Okay, let's move on to the next question. Okay, very similar looking question that we had taken for arithmetic progression, not taking it for the geometric progression now. So the question goes like this, if n is the set of all natural numbers and it is partition into groups like this, s1 is 1, s2 is 2, 3, s3 is 4, 5, 6, 7, s4 is 8, 9, 10, 11, 12, 13, 14, 15 and so on. Then find the sum of the numbers in s50, find the sum of the numbers in s50. Please leave your answer in exponent form, no need to calculate it because there will be huge in number. Any idea how to proceed? Yeah, sure. Okay, let's wait for at least 3 people to answer then I will discuss this. Okay, see there may be multiple ways to do it. Okay, so you may find your own way to find the pattern. No skandahan. Oh my God Aditya, you don't have to find that in powers of 10 and all. Okay, anyways, let's see this pattern here. The pattern here is, so let me write few of the terms, 8, 9, okay. The pattern here is that the first term is always the set number minus 1 raised to the power of 2. So this is your first term which is 2 to the power of 0 which is 1. This is nothing but 2 to the power set number minus 1, correct. This is nothing but 2 to the power set number minus 1, correct. This is nothing but 2 to the power set number minus 1 which is 8, correct. In the same way I can say that once I reach s50, the first term will be 2 to the power 49. Do you agree with me or not? And from here onwards I have to keep adding a 1 till I reach 50 terms. Till I reach 50 terms. Yes or no, correct. So it's an AP only whose first term is 2 to the power 49. Common difference is going to be 1. Number of terms is going to be, oh by the way, number of terms are not 50. Number of terms are going to be 2 to the power of 49 again. Yes or no? Yes or no? See when it is s2 there are 2 terms, s3 there are 4 terms, s4 there are 8 terms. So s50 will be 2 to the power 49 terms. So n is 2 to the power 49. So let's use the formula of sum of an AP which is total number of terms by 2. 2 into first term plus n minus 1 into D. So if you write this it will become 2 to the power 48. Please note that I have just cancelled out a 2 from the numerator and denominator. This is again 2 to the power 49 and this, sorry, 2 into 2 to the power 49 and this is again 2 to the power 49 minus 1. So you can write this as 2 to the power 48. 3 into 2 to the power 49 minus 1. So this is going to be your answer. I'm sure you would have all expanded it and written it, that's fine. Unless and until your answer is deviating from here, it is correct. What happened? Aditya? Oh my God. You don't have to calculate it. You can't do that without a calculator. Is that clear? Okay, let's take on another question. Let's take this question. If x is given as 1 plus a plus a square plus a cube till infinity, y is given as 1 plus b plus b square plus b cube till infinity, you have to show that 1 plus ab plus a square b square plus a cube b cube till infinity is going to be xy by x plus y minus 1. Both n and b are between 0 and 1. Just type done once you're done with it because the result is only given to us. So you just have to prove it. So once you're done, please let me know. Done. Good. Okay. I'll give one more minute to those who are trying it. Please wrap it up in one minute. Okay, let's discuss this. So basically here, what has been given to you is that x is 1 by 1 minus a, right? Because it's an infinite GP, right? So here the first term is 1 and the common ratio is a itself. So for this, it will be a by 1 minus r, which is this, right? Which means you can write 1 by x as 1 minus a, which means a is 1 minus 1 by x, which means it's x minus 1 by x. Okay. Similarly, y is 1 by 1 minus b, which means b is y minus 1 by y. Okay. Now look at the last series here. Last series sum would be nothing but 1 by 1 minus a b, which is nothing but 1 by 1 minus a b. You can replace your a with this and b with this. So it becomes x minus 1 by x into y minus 1 by y. Okay, just rearrange the terms. So xy you can take LCM and take it up. So it will be xy minus x minus 1 y minus 1. Okay. And if you expand this denominator term, it will be xy minus xy minus x minus y plus 1. Correct? Which will be xy. If you open the bracket, it becomes x plus y minus 1, which is actually your right-hand side. Okay. And hence proved. Simple. So the trick was to write your a and b in terms of x and y over here. Is that fine? Can we move on? Can we move on to the next question? Yes sir. Okay. Let's take this question. So read the question here. If the continued product of three numbers in GP, it just means the product. So even if you drop this word continue, no problem. If the product of three numbers in a GP is 216 and the sum of their products in pairs is 156, then find the sum of the three numbers. That's correct. Correct Aditya. That's correct. That's correct. Okay. Let's discuss this. So let's say the three terms are a by r, a and a r. Okay. So first of all, you have been given that the product of these three guys is going to be 216. Okay. Which means a cube is 216, which clearly means a has to be six. Correct. So your three terms are actually six by r, six and six r. Okay. Now what has been given to us in the second information that the sum of their products in pairs is 156. So in pairs, if you do six by r into six, six into six r and six by r into six r, that's going to be 156. By the way, the last term here is 36 by r plus 36 r is plus 36 is equal to 156. Correct. Which means 36 by r plus 36 r is equal to 120. Divide by 36 throughout. So one by r plus r is equal to 120 by 36. 120 by 36 is 10 by three. Correct. Now you can either convert it to a quadratic and solve for it or you can even guess that 10 by three is like one by three plus three, right? Which clearly implies that your r could either be three or r could be one by three. Correct. So if you take r as three, then your given series will be 2618. But the sum is going to be 26. If your r is one-third, your given series will be 1862. But nevertheless, the sum doesn't change. It is still 26. So your answer is going to be 26. Is that fine? Correct. Ronak, Shomik, Kirtan, Siddhartha, Veldhan. Okay. So here we'll take a five minute break. Okay. Let's take a five minute break here and then we'll resume with more problems on the other side. We'll be back at 6.38. So we'll resume at 6.38 p.m. So take a five minute break. The class is only till 7.30. Okay guys, welcome back. So hope you all are back after the short break. So we'll take another problem from GP itself. Let's take this one. So find a three digit number whose consecutive digits form a GP. If we subtract 792 from this number, we get a number consisting of the same digits written in the reverse order. Now if we increase the second digit of the required number by 2, the resulting digits will form an AP. So what is that number? Start working on this and reply to me privately. Anyone? Sir, one second. Yeah, yeah. Okay. That's correct, Ananya. That's correct. Correct Aditya, correct Kirtana. I need two more responses before I start solving this. Okay, let's discuss this. See, let's say the three numbers, the three digits of this number are A, AR and AR square. Okay. Now what is the first condition given to us? If we subtract 792 from this number, we get the number consisting of the same digits written in reverse order. Now remember the digit here is A, AR and AR square. So basically the number itself is 100 A, 10 AR and AR square. Okay. If you subtract 792 from it, you get the same thing in reverse order. That means 100 AR square, 10 AR plus A. Correct. So let me simplify this. So this is going to give you let me club 99 A from here. So this and this will be 99 A minus 792 is equal to 99 AR square. Correct. Now drop the factor of 99 throughout. It will become A minus 8 is equal to AR square. Right. Which means A1 minus R square is equal to 8. So you can write this as A1 minus R. Okay. Into 1 plus R is equal to 8. Now guys one more thing I want to highlight here is that the moment you write this, the game is over actually. How? Because you know that this is a positive term. So your A can only be 9. Are you getting my point? So there's no other possibility than A being equal to 9. Right. Because if you take any other term, let's say if you take A as 8, that means AR square is 0. That means either A is 0 or R is 0, but both are not possible. Correct. And let's say if you take a 7, you start getting a negative answer, which is not possible. Right. So if you're a very keen observer, your problem will be solved at this step itself. You don't have to do anything further. Correct. So if A is 9, you can say 9 minus 8 is equal to 9 R square. That means R is equal to one third. Correct. If R is one third and this is 9, this has to be 3 and this has to be 1. Over. Problem is finished here. But let us say if you want to extend this problem and solve it in a rigorous way, then how do we solve it? So that's what I'm going to do next. So from this step we got this. So we can call this as let's say 1. Okay. So the second condition given to us is absolutely waste. It is not required at all. So you could solve this with lesser information actually. Now, if we increase the second digit by 2, the resulting numbers are in AP. So basically what you are saying to say is that A, AR plus 2 and AR square are in AP. What does it mean? It means twice of the middle term is equal to the sum of the extremes A and AR square. Correct. So from here I can say 2 AR plus 4 is equal to A plus AR square, which means AR square. Let me take it on the other side. So can I say AR square minus 2 AR plus A is equal to 4, which is nothing but if you take an A common, it will be R minus 1 square is equal to 4. Let me call it as 2. Now using 1 and 2, I'm going to solve this. Okay. So I'll divide my 1 by 2. So what do I get? A1 minus R square divided by AR minus 1 whole square is 4 by 8. Is that fine? A gets cancelled. Numerator, I can write it as R minus 1, R plus 1. And this also I can write it as 1 minus R, 1 minus R. So that's equal to half. Let's cross multiply. So 2 plus 2 R is equal to 1 minus R. So 3R is equal to negative 1. Sorry, 3R is equal to... Oh, one second. Why am I getting negative 1? Somewhere we missed out anything. Just check. Oh, it's the other way around. 8 by 4. Yeah, sorry. So this is 8 by 4. So that's actually 2. Yeah, sorry. So this gives you 1 plus R is equal to 2 minus 2R. That means 3R is equal to 1. So R is equal to 1 third. So if R is 1 third, you can use any one of the expression to get your value of A. For example, you can put here. So A, 1 minus 1 third, 1 plus 1 third is equal to 8. So A into 2 by 3 into 4 by 3 is equal to 8. That means A is equal to 9. So your A is 9, R is 1 third, which clearly means this is 9, 3, 1, which we have already guessed earlier also. Is that clear everyone? This was an easy question. You should have done it. Can I move on to the next question? Okay. Now I'm sure in the school you would have done this sum. What is the sum of 7 plus 77 plus 777 plus 77777 all the way till, let's say, n terms? Let me know once you're done. No, that's not right, Mayur. No, Shankin. That's correct, Akshay. Okay. Guys, first of all, if you think that it's a GP, please let me tell you it's not a GP. Okay. So how do we sum this up? In fact, any problem of this nature where you have A, A, A, A, A, A, A, A, like that, and you want to sum this up, the way is first take the A out common. That means you take the 7 out common. So you end up getting 1 plus 11 plus 111 like that till n terms. Okay. Then multiply and divide with a 9. So it becomes 99, 999, 999, 9999 and so on till n terms. Okay. Now, this 9 can be written as 10 minus 1. 99 could be written as 100 minus 1. 999 could be written as 1000 minus 1 and so on. Till the last term could be written as 10 to the power n minus 1. Okay. Collect all the ones together. Okay. So you'll end up getting 10, 100,000 and so on till 10 to the power n. And if you collect all the ones, it'll be n. Correct? So this will be a GP with the first term being 10 and common ratio is also 10 and number of terms is also n. So you can write this as a r to the power n minus 1 by r minus 1 minus n. So that's going to be 7 by 9, 10 into 10 to the power n minus 1 by 9 minus n. Take an LCM of 9 inside the brackets. So you'll get a 7 by 81 and you get 10, 10 to the power n minus 1 minus 9 in. Okay. Which you can further simplify if you want. 7 by 81, 10 to the power n plus 1 minus 9 and minus 10. So this is your answer. Sir, this will work for any number. Yeah. Yeah. It'll work for any series like that. But please be careful. The approach is the same. Why divide by 9? See, what I'm doing is I'm trying to create a GP inside. Right. So what I did, I multiplied and divided both by 9. So it's not like I've changed the expression. So I'm dividing it also and multiplying it also. So that 9 could be treated as 10 minus 1. 99 could be written as 100 minus 1. 999 could be written as 1000 minus 1. So that 10, 100, 1000, etc. They will form a GP. That's the reason for multiplying and dividing by 9. Is that clear now? Okay. Let's move on to the next question. Okay. We'll move up a bit. Please read the question very carefully. If SP and R are the sum product and sum of the reciprocals of N terms of an increasing GP, then S to the power N is equal to R to the power N into P to the power K, where K is which of the following options? Please choose from A, B, C, D and type it privately. In the GP which we consider, should we consider R greater than 1 or lesser than 1? See, increasing GP means R would be greater than 1. Okay, sir. That's correct, Shankin. No Rangaraths, that's not correct. Keetana, yes, that's correct. Correct Aditya. Correct Aditya, Manjunath also. Okay, let's discuss. So first of all, what is the S? Okay. S is going to be A times 1 minus R to the power N by 1 minus R. Correct. P, what is P? P is basically A A R A R square all the way till A R to the power N minus 1. Isn't it like saying A to the power N and R to the power 0 plus 1 plus 2 all the way till N minus 1? Okay. Which is nothing but A to the power N, R to the power N N minus 1 by 2. What is R? R is the sum of the reciprocals. So isn't this also a GP whose first term is 1 by A, right? Common ratio is 1 by R to the power 1 by R. Correct. So you can write it as this. Yes or no? Yes, sir. So if you simplify this, it becomes R to the power N minus 1 by A R to the power N minus 1 R minus 1. Check it out. So you can write this as 1 minus R to the power N, A R to the power N minus 1, 1 minus R also. Okay. Now let us divide S by R. See, how do I get that idea because here S by R, you can write this as S by R to the power N. Correct. So if I, you can write this as P to the power K, right? Yes or no? So let me divide S by R. So when you divide S by R, basically what are you doing? You're writing A 1 minus R to the power N by 1 minus R divided by 1 minus R to the power N into A R to the power N minus 1, 1 minus R. So this term, this term gets cancelled. This term and this term gets cancelled. So that gives you A square R to the power N minus 1. Correct. Yes or no? Correct. So if you raise it to the power of N, I get A to the power of 2N into R to the power of N, N minus 1. Correct. Isn't this like saying A, N, R to the power N, N minus 1 by 2 the whole square, which is nothing but P the whole square? Correct. So this term is actually P whole square. So your K has to be equal to 2, which means option number B is correct. Is that fine everyone? You want me to go down here? Please let me know if you're convinced of this. Then only I'll proceed further. Yes sir. Clear? Yes sir. Clear? Yes sir. Now we'll talk about properties of GP, properties of a geometric progression. First property, if A1, A2, A3, etc. till A and R in GP with a common ratio of R, with a common ratio of R. Please note that even if you multiply a non-zero constant K to each of these terms, it will still remain in GP with common ratio R itself, absolutely. Similarly, if you divide by a non-zero constant K throughout, it will still remain in GP with common ratio R. R. But what about if you add or subtract a constant from each of these terms? They won't be in GP. Correct? It's not in GP. Okay. What about if you raise it to a power of N? Yes. Or power of P, let me call it. N is the number of terms, right? Power of P. P being some rational number. Then this will still be in GP with common ratio. With common ratio R to the power P. Okay. This also includes minus one also, right? Because 1 by A1, 1 by A2, 1 by A3, etc. will also be in GP. Okay. What about log A1, log A2, log A3, etc. What will this be? This will be in AP with common difference as log of R. R being the common ratio. Are you getting this point? This is very important. Is that clear everyone? Okay. If your B1, B2, B3, etc. are also in GP with common ratio as let's say R2. Okay. Then please note that A1, B1, A2, B2, etc. will also be in GP. What will be the common ratio here? R1, R2. Yes. With common ratio as R1, R2. And same will go for even if you take their ratios. This will be in GP with common ratio as R1 by R2. Is that fine? Okay. Next property is any Rth term could be written as under root of terms which are equidistant from R. Now this is a very important property because it's going to help you to understand what is the geometric mean. Okay. So if A, B, C, R in GP, please note B is under root of AC or that means B2 is equal to AC. Okay. You can easily prove this. You can easily prove this. What is AR? AR is A, R to the power. Okay. Let me not write R here because R has only been used. Let me write it as P, P to the power R minus 1. AR minus K will be what? A, P to the power R minus K minus 1. AR plus K will be what? AP to the power R plus K minus 1. Correct. If you take the product of these two, let me write it here. If I multiply these two, I get AR plus K into AR minus K as A square and please realize that you'll get P to the power 2 times R minus 1. So if you take an under root of it, that means it will become AP to the power R minus 1 which is nothing but your AR itself. Are these properties clear to everyone? Yes, sir. Okay. So well, without much waste of time, we can start talking about harmonic progression. I'm sure this must not have been done in the school. But does anybody know what's in harmonic progression? So reciprocal of an AP series. Absolutely. Absolutely. Sir, sorry, can you go to the previous slide one second? Sure. No need to copy it. I'll share it with you. I'll share the PDF with you guys. Don't worry. Thank you, sir. So thank you, sir. Okay. Yes, you are absolutely right. A harmonic progression is nothing but it's a reciprocated version of an arithmetic progression. So basically, if you reciprocate all the terms of an arithmetic progression, you end up getting a harmonic progression. Are you getting it? That's fine. Most of the cases when we are given a problem on harmonic progression, what do we do is we reciprocate all the terms of harmonic progression and start treating it as an arithmetic progression. Are you getting my point? So a few examples I can cite over here. Let's say 1 by 1, 1 by 2, 1 by 3, 1 by 4, etc. They are in HP. Right? Because if you reciprocate these terms, you'll get 1, 2, 3, 4, which is in AP. Okay. So the biggest advantage here is that you can use all your understanding of an AP to solve questions of an HP. So HP is nothing but all the reciprocated versions of AP. Okay. Now, good or bad, there is no direct formula for getting the sum of an HP. Okay. So HP, there is no direct formula to get the sum. Okay. But one thing very interesting which I would like to highlight over here, important finding, which is useful for your exam also. If A, B, C are in AP, please note 2B is equal to A plus C, which also makes you rewrite this as A minus B by B minus C is equal to A by A. No, no, no, no, Akshar, we can't do that. Life is not that simple. If you find the sum of an AP and just reciprocate it, we won't get the sum of an HP. Sorry, it doesn't happen. It's like saying 1 by A plus B is equal to 1 by A plus 1 by B, right? Which is wrong, right? Absolutely wrong. We can't do that. Unfortunately, we cannot do that. Okay. If A, B, C are in GP, it means B square is equal to AC, which also means A minus B by B minus C is actually A by B. I think this I already told you, but this, I think I'm telling you right now. And finally, if A, B, C are in HP, please note B is equal to 2AC by A plus C. Can you prove this first? Can you prove that if A, B, C are in HP, B will be 2AC by A plus C? And further, please, you can say that A minus B by B minus C is equal to A by C. Can you prove this quickly? Just type done if you're done. Okay, doesn't take much of a time. Just saying A, B, C are in HP, indirectly you're saying 1 by A, 1 by B, 1 by C are in AP, correct? What does it mean? It means double of the middle term. So 2 into 1 by B is equal to the sum of the extremes, which is 1 by A and 1 by C. So let us simplify this. It becomes A plus C by AC, which means B is equal to 2AC by A plus C. Okay. So this further implies that A minus B by B minus C is equal to A by C. Now, why I gave you these three results separately was the reason because this has been asked as a direct question in one of the exams. Is that clear? Okay. Let's take a question on HP first. Question is, if the sum of three terms in HP, if the sum of three numbers in HP is 37 and the sum of their reciprocals and the sum of their reciprocals is 1 by 4, we write it properly else. Okay. Then find the numbers. Please type it out privately to me. What are the numbers? No, that's not correct. Siddhartha, that's correct. Shomik, no Aditya, no, no, no, no, no, no, no. You have done a very, very silly error here. The answer is whatever you have written in a reciprocated way. So only Shomik has given the right answer so far. That's correct, Ronak. Well done. No, Trippan, that's not that ugly. Yeah, Aditya, no, that's correct. Good. Guys, one more minute for the people who are wrapping it up. Still not correct, Siddhartha. No, Ananya, that's not correct. No, Aditya. Yes, Siddhartha. Middle term is now you're correct. No, no, no. Okay. End of time has been given to this. All sorts of answers I'm getting. See, if you have to take three terms in an HP, better to take it as 1 by A minus D, 1 by A and 1 by A plus D. Right? So this will be the three terms in HP. Okay. So first information is that the sum of these terms is given to you as 37. This is 37. Correct? And the second set of information is that if you reciprocate these terms, it's 1 by 4. Correct? So from this, I can say 3A is 1 by 4. So A is equal to 1 by 12. Correct? Now, let's use here. So 1 by A minus D. In fact, let me write A itself as 1 by 12. So 1 by 12 minus D plus 1 by 1 by 12 is 12. And 1 by 1 by 12 plus D is equal to 37. Correct? Which means 12 by 1 minus 12 D plus 12 by 1 plus 12 D is equal to 25. Right? So I can write it as 1 by 1 minus 12 D and 1 by 1 plus 12 D is equal to 25 by 12. Take the LCM, cross-model applies. So you'll end up getting 2 by 1 minus 144 D square. Correct? That's equal to 25 by 12. Okay? Bring these two down and make it 24. Just reciprocate it. This will be 24 by 25. Which means 144 D square is 1 minus 24 by 25 which is 1 by 25. Correct? Which means D square is 1 by 44 into 25. Is that clear? Which happens to be 1 by 12 square into 5 square. So D is equal to 1 by 60 or minus 1 by 60. Is that correct, guys? Okay. Now, let us wrap up our series here. So if you see here, it will become 1 by 1 by 12 minus 1 by 60. I'm assuming D to be 1 by 60 as of now. Okay? This term will be 1 by 1 by 12. This term will be 1 by 1 by 12 plus 1 by 60. Which gives you... Let us simplify this further. This is 1 by 5 by 60 minus 1 by 60 which is 4 by 60. This is already 12. And this is 1 by 6 by 60. So your terms will be 15, 12 and 10. Getting the point? If you take D as minus 1 over 60, then let's see what happens to these terms. So this first term will become 1 by 1 by 12 plus 1 by 60. Next term will be 1 by 1 by 12. And the next to next term will be 1 by 1 by 12 minus 1 by 60. Which actually becomes the same terms in reverse fashion. So 10, 12 or 15. So your three numbers are either 15, 12 or 10 or 10, 12 or 15. Is that clear? I don't know what kind of mistakes you were making. I'm sure there were all series mistakes. So, well guys, we'll wrap this up now. Oh, Mac was Irish. I came to know just now. By the way, before we meet for the next class, here are the topics that I'll complete for you. I'll be talking about different types of means for you. Which includes AM, GM, HM. I'll also talk about series. So basically we'll talk about special series. We'll also talk about how to sum up a series by the use of methods of summation. Which primarily includes two methods. One is called the sigma method of summation. And the other one is called the method of difference. Okay. Under method of difference, there's a method which is VN method. We are going to spend a lot of time on that. Okay. So this will be our agenda for the next class. If we have time permits, we'll be completing even our binomial theorem for non or you can say negative or fractional powers. Okay. You can say non integral powers. Non positive integral powers. Okay. So guys, thank you so much. Bye bye. Thank you sir. Thank you sir. Thank you. Thank you sir. Thank you everyone.