 Hello, welcome all to the YouTube live session on progression and series So those who have joined in now I would request you to type in your names in the live chat box so that I know who all are attending this session Hi, sure. Good morning. So good morning to everyone who has joined the session request a request you to type in your names in the chat box So we are going to start with this chapter, which is a very very simple part of ebra class 11 topic progression and series So this would be a first class on this. So we all know that sequence progression and series There's a difference between these three terms right What's the sequence sequence is basically a listing of Terms where they have a certain pattern or a rule governing the listing, right? But that rule may not be formalized that means you may not be able to express it as a formula for example For example The sequence of prime numbers the sequence of prime numbers There's no formula for finding out a prime number when I give you a position of it For example, if I say tell me the 78th prime number, there's no formula as such to find out the 78th prime number, but there is a rule. There is a pattern which governs the listing of prime numbers, right? In the same way Fibonacci sequence So there's no formula as such but there's a rule definitely that means every Succeeding term is the sum of the preceding two Okay, barring the first two terms So what is progression progression is nothing, but it is such sequences Where you have a formula for Tn that means the nth term could be obtained by a Formula that means if you provide the position of n that is the position of the term and Being some natural number then you will always be able to find out the value of the term, right? On the other hand series is basically nothing but the sum of terms of a sequences Sum of terms of sequences, okay? That's going to be a series. So don't Interchange these words. They have different meaning. So from this conclusion So from this discussion, you can draw a conclusion that progressions are sequences. So basically, let's say sequences is a bigger set like this Okay, and progression could be called as a subset of Sequences so progressions are nothing but sequences where you can formalize the The bad the position value of a term when its position is known Okay, so we are going to begin this course with the discussion on progressions So in this chapter, we are going to study primarily three types of progression the first being arithmetic progression Something which you have already studied in the junior classes in class 10th Then we are going to learn something new which is the geometric progression and we are going to touch upon Harmonic progression and finally we are going to talk about arithmetic or geometric progression arithmetic or geometric progression So without much ado, I'm going to start with arithmetic progression. So what's an arithmetic progression? You have already learned this So arithmetic progression is that sequence arithmetic progression is that sequence where Every term differs from its preceding By a fixed difference or by a constant difference So if I write terms like this a A plus D a plus 2d a plus 3d Etc. This would be called as an arithmetic progression because every subsequent term differs from the preceding by a fixed Difference D which we call as the common difference Okay, now this common difference could be zero as well This common difference could be positive this common difference could be negative and not only that this common difference could also be Imaginary so this can be zero This can be positive. This can be negative and this can also be imaginary So we can also have imaginary common differences, okay A is called the first term. So first term is actually a in this case Now if you go by this trend if you go by this trend What would be the nth term of this particular progression that is something which we all know This is your first term. This is your second term. This is your third term. This is your fourth term If I continue so on your nth term is going to be a plus and minus 1d Right, so this is something which is known to all of you not a new thing for you So on the basis of the nth term, let me start with a question Let me start with a question on the nth term. Yeah, my question is if if The mth term of an AP if the mth term of an AP is 1 by n and nth term is 1 by M Then Then find M nth M nth term of the AP So guys, please solve this and post your response in the live chat box By the way, good morning to everybody Shreya, Harima and Akash, Gaurav, Yamini, Gargit, leave welcome to the session So please post in your response once you have done. I'm sorry Absolutely correct to leave answer is one should be a very simple problem actually correct Shreya. That's absolutely correct So basically we can write this as a plus m minus 1 d is 1 by n and And a plus n minus 1 d is equal to 1 by m Okay So let us subtract these two when I subtract these two I get m minus n times d is equal to 1 by n minus 1 by Which means d is equal to m minus 1 and d is equal to m minus n by mn so d is equal to 1 by mn d is equal to 1 by mn and If I put the value of d in any one of the terms If I put the value of d in any one of the terms, let's say I put over here. I Get a plus m minus 1 by mn is equal to 1 by n Which means a plus 1 by n minus 1 by m is equal to 1 by n Which implies a is equal to 1 by mn right So 1 by mn plus Mn minus 1 into 1 by mn is going to be your m nth term So when you expand it, it's going to be 1 by mn plus 1 minus 1 by mn That is going to be t m nth term, which is actually going to be one So the answer for this is going to be one. Is that fine super simple question? All right next question Next question consider two APs Consider two APs Let's say I call AP 1 and AP 2 Whose terms go like this? two seven two seven 12 17 and so on till 500 terms and so on till 500 terms And there's another AP which goes like this one eight 15 22 and so on till 300 terms Till 300 terms Okay Question is find the last Common of these two APs of AP 1 and AP 2 Find the last common term of AP 1 and AP 2. Yes, anybody any response? Gargi, Shreya, Trideeb, Yamini, Gaurav, Arimaan Akarsh Anyone so let us first find out what is the 500th term of AP 1 so for AP 1 let us find out what is so 500 term would be 2 plus 499 times 5 Which is going to be 2497 correct Similarly for AP 2 Let's find out the 300th term. So it's 1 plus 299 times 7 Which is going to be 2100 plus Minus 6 will be 2094 Okay, so it'll be 2094 Now clearly a 2094 is smaller than this right So my last common term has to be less than equal to this term So by the way, let us see how many terms are common between these two AP So when I list them down to 7 12 17 22 27 32 37 42 47 52 57 62 like that list few terms here also a list 1 8 15 22 29 36 43 50 So I can see first common term is 22 in both the places second common term is 57 in both the places now once you've got these two terms We can conclude that the common terms will follow a AP like this so the common terms Will follow an AP like this 22 57 next term will be 92 like this Next term will be 92 like this. So basically it's having a common difference of LCM of 5 and 7 note that D1 for this is 5 D2 for this is 7 So the common difference over here will be LCM of 5 and 7 That's going to be 35 Okay So let us say nth term is the last term. So let me write this as less than equal to 2094 So let us simplify this so 35 times n minus 1 is less than equal to 20 7 2 okay, and 2072 divided by 35 will give me 59 point 2 so n minus 1 is less than 59 point 2 So n is less than 60 point 2 That means n is equal to 60 because n can only be a natural number because it represents the position of the term Right so 60th term of this particular AP would be your Last common term so 60th term will be this into 59 into 35 That's going to be two zero six two two zero six two will be your last common term Okay Just once again, let me check if Question is correct. Oh, sorry two zero eight seven three div is correct. Yeah, two zero eight seven. Yeah So two zero eight seven would be your Last common term no, I know Shreya because the common terms and You have to take n in one then you have to take m in the other You cannot have the same term equal in both the term So what you're trying to say is that your nth term of both the APs are equal which is not possible for example here the Fifth term is equal to the fourth term Right, so you cannot use n throughout you have used nn in both the places You have to change it as n in one place and m in the other Okay Now moving on to the next concept which is the selection of terms in an AP selection of terms In an AP now if you have to select three terms if you have to select three terms in an AP Then preferably we select it like this Especially when information related to their sum is given to us. Okay. Now. I'm saying preferably Preferably it is not mandatory Preferably you have to select it like this Not mandatory. Okay Similarly if you're if you ask to select four terms in an AP preferably we select it like this Especially when information related to their sum is given to us If you're asked to select five terms we select it like this a minus 2d a minus d a a plus d a plus 2d Similarly 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 now Why do we select terms like this because you would realize that Whenever you are adding them, let's say when you add all the three terms over here D will vanish So on adding D will vanish Okay, so you can directly find one of the parameters a over here Okay, no, it's preferably please do not think that you have to select it like this Now remember a is no longer your first term. Okay So here the first term is a minus d here your first term is a minus 3d like that Okay, and in some cases you'll find that the common difference is not D common difference is 2d actually So for example in fourth term the common difference is actually 2d Similarly in sixth term also the common difference is 2d and so on So, uh, let us take a quick question on this our question is if the sum of three terms if the sum of three numbers in ap is 24 and their product Is 440 find the numbers 5 8 11 or 11 5 absolutely correct Absolutely correct. So basically it's a very simple problem So you can choose the numbers to be a minus d a and a plus d So when you add it you get 3a is equal to 24. So a is going to be 8 And the product is 440 means a minus d 8 8 plus d is equal to 440 That means 64 minus d square is equal to 40 440 by 8 which is going to be 55 So d square is going to be 9 in this case so d can be plus or minus 3 So if you take plus and minus 3 So if you take d s 3 your terms will be 5 8 11 And if you take d s minus 3 You your terms will be 11 8 5 simple problem next one divide divide 32 into four parts Divide 32 into four parts Which are in ap Such that Such that the ratio of The product of the extremes the product of the extremes To the product of the means to the product of the means Is 7 is to 15 Is 7 is to 15 Hope you understand the meaning of extremes Extremes are the last two means are the middle two absolutely correct Absolutely correct. So the division is 2 6 10 and 14 akash siddhi absolutely correct guys It's simple first of all if there are four terms we can take the terms to be a minus 3d a minus d a plus d a plus 3d Okay, so let these be the four divisions of 32 of course their sum is equal to 32 So 4 a should be equal to 32 Means a is 8 Okay, now the product of the extreme twos Will be a square minus 9d square divided by product of the means will be a square minus d square This is 7 by 15 Which implies 64 minus 9d square by 64 minus d square is 7 upon 15 Okay And if you cross multiply you get 64 times 15 minus 15 times 9 which is going to be 135 d square is 64 into 7 minus 7d square right So I get 128 d square as 64 times 8 Okay, so this will be getting cancelled off by a factor of 2 this will get cancelled by a factor of 2 again So I'll have ds plus minus 2 So my terms would be 8 minus 6 which is 2 8 minus 2 which is 6 8 plus 2 which is 10 And 8 plus 6 which is going to be 14. So this will be your division This will be your division of Of the term 32 Okay Moving on to the next concept which is the sum of n terms of an ap sum of n terms Of an ap Now before I deduce that I'll tell you a small story which happened in germany 200 years ago There was a small child Six years old He was given a problem to add all the natural numbers from 1 to 100 Okay as a punishment by His teacher. He was just six years old. So the teacher was expecting that this Small boy will actually sit and count these numbers and it would take him a long time to count So it was actually a punishment given to him Now this guy He went home and he wrote this sum like this And he just reversed this entire series So he started writing it upside down for example 100 he wrote as the first term then 99 then 98 and so on Like this Then what he did he added these two series. So s plus s will be 2 s 1 plus 100 will be 101 2 plus 99. So he started adding these two terms 2 plus 99, which is again 101 Again, he added these two terms, which is again 101 He kept on doing that and he realized that all the terms actually get added to 101 Correct It made very simple for him to actually find the sum of all hundred and ones written 100 number of times and he found out the sum to be Very very easily as 5050. Okay This fellow went on to become the famous mathematician and physicist of his time And you would be knowing his name. His name was Carl Frederick Goss Carl Frederick Goss You must have learned about the goss law in physics. So this was a small child who was given this punishment Now taking a clue from what this small child did we actually can find out the sum of n terms Of any generalized arithmetic progression And how do we do that? So let us say the series is a a plus d A plus 2d and till nth term it will become a plus n minus 1d And we can write the same in a reverse fashion like this And all the way till a Now add these two series when you add it becomes 2s Now some of these two will become 2a plus n minus 1d Here also when you add these two it becomes 2a plus n minus 1d And if you keep on doing it even till the last term when you add these two will get 2a plus n minus 1d So in other words, it's nothing but 2a plus n minus 1d written n number of times. So this will become your required sum So s will become n by 2 Into 2a plus n minus 1d Okay Now what we learned from this experience is that The sum of terms equidistant from the beginning and the end will always add up to give you the same number, isn't it? So what we concluded here the conclusion is the sum of terms sum of terms equidistant equidistant from Start and end Is a constant and that constant is actually the sum of the first and the last term sum of the first and the last term yes or no So whatever the sum of the first and the last term same is the sum of the second and the second last Same is the sum of third and the third last This is actually a very important property of an arithmetic progression any arithmetic progression if you take You realize that some of the terms equidistant from the beginning and the end would be the same for example, if I take one three five seven nine 11 you realize that 1 plus 11 is same as 3 plus 9 will be same as 2 plus 5 And if there happens to be a middle term Then it would be double of that middle term. That means the middle term added to itself. So example 1 plus 13 is 14 7 3 plus 11 is also 14 5 plus 9 is also 14 and if there's a middle term it will be 7 plus 7 which is also 14 okay So in light of this particular conclusion, we can also claim that the sum of n terms of an arithmetic progression could actually be written as sum of the first and the last term times n by 2 So if I know the total number of terms if I know the first term and the last term I can always write it like this I can also write it as sum of the Second and the second last term I can also write it as the sum of the third and the third last term like this So I can also write it as I can also write it as sum of the kth and k n minus k minus 1 right And if there happens to be a middle term if the middle term is Let's say t m Then the sum of n terms could be n times the middle term. Please remember that It will be n times the middle term Guys, can you all see this screen? So let me begin with a problem on this concept. The first problem that we're going to solve is if a1 a2 a3 etc Till a24 are in ap Okay such that such that a1 plus a5 a1 plus a5 plus a10 plus a15 plus a20 Plus a24 is known to us as 225 find Sum of 24 terms of this arithmetic progression find the sum of 24 terms of this arithmetic progression our answer is 900 Okay, what about others? Anybody else who thinks the answer is 900? So guys just now I told you in any arithmetic progression The sum of the first and the last so t1 and t24 is the first and the last term. Okay, so let's say I call this as x Okay, so this is your fifth and the fifth last term. So this will also be x This is your 10th and the 10th last term. So this will also be x. So basically what is given to us is 3x is equal to 225 So x is given to us as 75 Okay, and remember The sum of n terms will be n by 2 the sum of the first and the last term So their total number of terms is 24. So 24 by 2 some of the first and the last term is going to be uh x Which is 75. So it's 12 into 75. That's going to be 900 absolutely correct rr man. So rr man was the first person to answer this next question is next question is if the ratio of the sum of if the ratio of the sum of two ap's of n terms is 5n plus 4 divided by 9n plus 6 Then find the ratio of their 18th terms find the ratio of their 18th terms Uh, in short, I will write this. What does it mean? It means that you have been given that Let's say s n is the sum of the first ap till n terms and s dash n is the sum of second ap to n terms This ratio is given to us like this What I have to find is t 18 by t dash 18 Okay, so shia says 127 upon 321 Anybody else who has a similar answer or another answer? 3d says 179 is to 321 Okay So guys, uh, I need one more response Then I'll start solving it. Okay. Shia. Shia has decided to correct her answer. Okay. No problem. Shia happens Gargi, Gaurav, Aariman Akarsh Yamini It's nine. It's this is nine. All right. So s n by s dash n. I can write it as n by two Okay r 2a plus n minus 1d Similarly for the second ap I can write it as 2a dash plus n minus 1d dash And this ratio is 5n plus 4 by 9n plus 6 n by 2 n by 2 cancels Okay Now what you can do is you can Divide the left hand side of the term by 2 So if you divide it by 2 you get n minus 1 by 2d divided by a dash plus n minus 1 by 2d dash This still remains 5n plus 4 by 9 and plus 6 Now what do I need? I need the ratio of the 18th term of both the progressions which is actually a plus 17 d by a dash plus 70 In other words, if I somehow able to put 17 over here My job will be done because I'll be getting the ratio of the 18th term So what I will do is I'll substitute n minus 1 by 2 to be 17 n minus 1 by 2 to be 17 Which means n is 35 So let me substitute that in this equation put n as 35 You get a plus 17 d by a dash plus 17 d dash Is 5 into 35 plus 4 by 9 into 35 plus 6 which happens to be 179 by 321 so 3 for the first one to answer this correctly Well done. By the way, this type of problems you have done in your class 10th exams In your class 10th Very good. All right. So moving on to the next question Next question goes like this if if n natural numbers if n natural numbers Is partitioned Is partitioned into groups Like this s1 is 1 s2 is 2 3 s3 is 4 5 6 s4 is 7 8 9 10 et cetera okay then find the sum of numbers in s 50 Then find the sum of numbers in s 50 3d that would not be a right approach because their ratio is given to you So ratio may end up cancelling the common factors So exactly you cannot say that the coefficients will be your first and the The constant term will be your first term and the ratio of n will be a common difference It's like saying that If tan theta is equal to 2 by 4, then you're trying to say sin theta is 2 and cos theta is 4 That's absurd, right? Okay, so you cannot directly do that in here you were lucky that you got the answer And probably in this case it did not matter But we do that absolutely carefully Kargi says 4 8 5 1 All I can say kargi is your answer is wrong Try again. It's much more than that Shia says a 6 2 5 2 5. Okay. Shia. I acknowledge your answer Let us wait for other response Okay, so aria man is also of the same opinion 6 2 5 2 5 I need one more response All right. So akash also thinks the same. So guys, how do we do this problem? I want you all to notice one thing that if you see the last term So last term of s 1 is 1 Okay Last term of s 2 is 3 which is actually 1 plus 2 Right last term of s 3 is 6 which is actually 1 plus 2 plus 3 Right last term of s 4 is 10 which is actually 1 plus 2 plus 3 plus 4, right If I go on doing this, can I say the last term? The last term of s 50 will be 1 plus 2 plus 3 all the way till 50 Right, which is nothing but 50 into 51 by 2 That's going to be 51 into 25 Right, which is actually 1 2 7 5 1 2 7 5 right now Think as if you're going back now. This is 1 2 7 5 then 1 2 7 4 and so on I'm going back to whichever term I don't know but I'm going back by 50 terms So I'm trying to sum up all these terms I'm going to sum up all these terms for my s 50 sum right So in short, I'm doing The sum of terms of s 50 as 1 2 7 5 1 2 7 4 And so on till 50 terms So I can write this as The formula which I know s sum will be n by 2 2 times the first term Plus n minus 1 the common difference is minus 1 now so that's going to be 25 and This will become 2550 And this will be minus 49 So it's 25 into 2501 So 25 into 2501 that will give you 6 25 25 So the first one to answer this was Shreya Shreya well done very good. Yeah, absolutely Gaurav Absolutely correct Next problem Now the problems will become slightly difficult now. Okay next problem is if A B C D are distinct integers Are distinct integers and they form an increasing They form an increasing arithmetic progression increasing arithmetic progression means subsequent terms are bigger than the preceding Okay such that such that A square plus B square plus C square is actually D Okay Then find the value of Find the value of A plus B plus C plus D Value means a constant will come don't express it in terms of A B and C So find the values of A B C and D. So they are distinct integers. So watch out for these words They are distinct integers form an increasing AP Then find the value of A plus B plus C plus D Because here you have to play on the fact that these terms are integers and they form an increasing AP Okay, so let's let's do this problem now since D is already used I can use the fact that B is a plus. Let's say common differences capital D C is a plus 2d And let's say D is a plus 3d Now let's use this condition that a square plus B square plus C square is small d So a square A plus D the whole square A plus 2d the whole square is equal to a plus 3d Okay, so let us simplify this So when you simplify this all together, we will have around uh, how many d squares 5d squares Then you'll have 2 ad and 4 ad which is uh 6 ad 6 ad And minus 3d So 6 ad minus 3d And the constant terms would be uh, 3a square coming from this and a coming from this. So this would be 0 In other words, I get something like this Okay Now this is a quadratic equation. This is a quadratic equation in capital D And I'm sure that D that comes out from there should be real and distinct Right because we have a real AP formed over here So can I say discriminant? Discriminant Should be greater than equal to zero Okay, so discriminant would be B square Minus 4 ac This should be greater than equal to zero. So I can take a factor of uh 3 out so 9 times 2a minus 1 the whole square minus 20 3a square minus a is greater than equal to zero So let me simplify this so 9 times 4a square minus 4a plus 1 Minus 60 a square plus 20 a should be greater than equal to zero So 36 minus 24 is 36 minus 60 is 24 minus 24 a square minus 36 This will become minus 16 a Okay Plus 9 This should be greater than equal to zero That means 24 a square plus 16 a minus 9 Is less than equal to zero Yes, absolutely correct. The answer is 2 actually Now let me find the roots of this. Let me find the roots alpha and beta for this So let's say alpha and beta are the roots of this quadratic in a Okay, so I'm trying to solve this inequality. So let's find alpha and beta alpha beta would be nothing but minus a Sorry minus 16 minus b Plus minus b square under root b square minus 4 ac Under root b square minus 4 ac which will be 256 minus In fact will be plus 4 into 24 into 9 by 48 by 48 Okay, so b square minus 4 ac will be this term So let me simplify this so 256 plus This will become 1 1 2 0 So minus 16 plus minus under root 1 1 2 0 By 48 So I can write this as minus 16 plus minus 4 under root 70 By 48 which is minus one third Plus root 70 by 12 Okay, so the two roots will be this So I know that here a should lie between alpha and beta When such an inequality comes we have all learned in our linear inequalities that a should happen to lie or the variable should happen to lie between the roots So it should lie between alpha and beta So between alpha and beta means it should lie between minus one third minus root 70 by 12 and minus one third plus root 70 by 12 Okay Now the approximate value of this is minus 1.03 Okay, and for this it is 0.036 approximately Now a has to be an integer a has to belong to an integer So the possible values of a which is between these two numbers is a could be minus 1 or a could be 0 Okay Now Having found the value of a as minus 1 and a as 0 Let's try to figure out what happens when a is 0 when a is 0. What will be the value of db? Okay So let's go on to find the value of d when a is 0 So I think we already had a quadratic over here Okay, so in this quadratic when I put a as 0 I get 5d square Minus 3d Plus 0 equal to 0 right So here d could be 0 or d could be 3 by 5 Now is this possible Is d equal to 0 possible? No because it's an increasing ap Is d equal to 3 by possible? No because in this case your d will not be an integer Remember d should also be an integer because all the entries are integers Okay, because all the entries are integers. We cannot have these two possibilities So a equal to 0 is ruled out a equal to 0 is ruled out Now let's take a as minus of 1 If I take a as minus of 1 let me again check what is the value of d that we get When a is minus 1 d becomes 5d square minus 9d And this will become 3 plus 1 which is 4 So this can be factorized as 5d square minus 5d minus 4d plus 4 equal to 0 Take 5d common So here d comes out to be 1 or d comes out to be 4 by 5 4 by 5 is again ruled out but 1 is possible but 1 is possible so My a is finally My a is finally 1 and d is Sorry a is minus 1 and d is 1 So a becomes minus 1 b becomes a plus d which is 0 c becomes a plus 2d which is going to be 1 And again d is going to be 2 So this is your a this is your b this is your c and this is your d So the sum of them is going to be Minus 1 plus 0 plus 1 plus 2 which is going to be 2 in this case. So 2 is your answer Yeah, I know many of you would have worked by guessing the values Okay, so this type of question will you can expect to come in kvpy type of exams They simply love these type of questions next question if a1 a2 a3 Till an are in ap Where all the ai's are positive Okay, then 1 by root a1 plus 1 by root a2 Plus 1 by root a2 plus 1 by root a3 And so on till 1 by root a minus 1 plus under root an Is which of the following options? It's which of the following options Is it 1 by under root a1 plus under root an is it 1 by Under root a1 minus under root an Is it n by Under root a1 minus under root an Or is it n minus 1 by under root a1 plus under root an So choose from the given four options a b c and d Gaurav says option number d How about others I need at least two more response To start solving this problem All right, so let's start solving this. So let's take the first term 1 by root a1 plus root a2 Whenever you see such a term the obvious feeling that comes is let us rationalize this So let us rationalize this by multiplying and dividing it with under root a1 minus under root a2 Should becomes under root minus under root a2 and the denominator I'll have a1 minus a2 Remember a1 minus a2 a2 minus a3 a3 minus a4 etc They will all be They will all be same as minus of d d be the common difference So I can write it as root a1 minus root a2 by minus of d Okay, so when I'm adding all these things can I say I'll start getting terms like this a1 minus root a2 plus root a2 minus root a3 And so on till root an minus 1 minus root an And you realize that terms will start getting cancelled Leaving you with minus 1 by d under root a1 minus under root an Okay Now I can simplify this by further rationalizing it My when I further rationalize it I get root a1 Minus root a1 times root a1 plus root an Divided by root a1 plus root an so this will give me minus 1 by d a1 minus an by Under root a1 plus under root an Now we know that an is a1 plus n minus 1d So n minus 1d is nothing but an minus a1 Which means I can write d as an my This divided by n minus 1 Okay, so minus 1 by d I can write it like this So I just substituted by d back I substituted this d over here So this and this goes with a minus sign and minus minus also gets cancelled leaving me with under root n minus 1 root a1 plus root an And that's what is your option number d saying so gaurav was the first one to answer this well done gaurav All right, so now we'll take up some properties We'll take up some properties of Arithmetic progression The first property is If a1 a2 a3 etc till an Are in ap with a common difference of d With a common difference of d Then note the following things Even when you add or subtract a constant from each of the terms, let's say a1 plus minus k or a2 plus minus k or a3 plus minus k And so on you still get an ap With no change in the common difference the common difference will still remain d Okay if you Multiply each term with a common difference. Sorry with a constant term k non-zero constant term k okay Then this will result into Again an ap Now in this case the common difference will now become k times d The common difference will become k times d k should not be zero in this case K should not be Zero in this case And a similar way if you divide by k Every term This will still be an ap With common difference being d by k Common difference will become d by k Okay If b1 b2 b3 etc are also in ap With a common difference of d Let's say this is d1 and this is let's say d dash Then remember the sum and differ or difference Of the corresponding terms Of these two ap's Will also be in ap With the new common difference as d plus minus d dash Plus minus d dash But if you multiply them That means a1 b1 a2 b2 a3 b3 etc they will not be In ap Or if you take the ratio also, they will also be not in ap Okay, if three terms are in ap Always remember 2b is equal to a plus c In fact, I will already explain this in a broader sense. That is The sum of the terms equidistant from the beginning and the end will be a constant And that would be equal to the first and the last term Is that fine? Now I have some questions For you based on this concept The first question is If abc are in ap If abc are in ap Prove that a square b plus c b square c plus a And c square a plus b are also in ap That's the first question Second question is prove that 1 by under root b under root plus under root c 1 by under root c plus under root a 1 by under root a plus under root b are also in ap and finally a times 1 by b plus 1 by c b times 1 by c plus 1 by a And c times 1 by a plus 1 by b Are also in ap. So in case you've done the first one, please send the snapshot of it on my personal whatsapp ID Yes, anyone any response All right, so let me start by the fact that The first three numbers over here, they are in ap. Let me consider that let me assume that they are in ap When they are in ap I can say that the difference of the Second and the first should be equal to the difference of the third and the second right So b square c plus a minus a square b plus c Should be equal to c square a plus b minus b square c plus a So let me do one thing. Let me take this two term and these two term together. So b square a minus a square c And the remaining term that is a b square minus a square b together And I will do the same term over here also c square a minus b square c term together and c square b minus c square b minus Or let me take b square a over here b square a and c square b minus b square c together, okay So this will become Take a c common So here also we can do the same thing actually on this side. We can take b square Yeah, in this side we can take b square c Yeah, b square c a square c a square c so b square c and a square c Okay, so b square c and a square c. So we'll have a b square minus a square b Yeah, so take a c common over here. It becomes b square minus a square take a b common b minus a Similarly, take an a common over here c square minus b square Take a b c common from here c minus b Now I can take b minus a common from here. So c b minus a b plus a a b b minus a here also a times c minus b c plus b And b c c minus b Take b Take b minus a common You get a b plus b c plus a c And here you take c minus b common c minus b common again you get a c plus a b plus b c So these two we can cancel off that means b minus a is equal to c minus b which means to b is equal to a plus c Which is actually true because a b c are in a b So what I did I I assumed this to be true and got to a condition which was already true That means my assumption was correct. This implies the assumption that a square b plus c b square c plus a And c square a plus b are in ap Is correct. Let me tell you this is an ncrt question This is an ncrt question. So sometimes ncrt also gives good questions All right, second one anybody could solve it So this is done first one is done second one anybody could solve it Again, you can take a similar approach. You may assume that 1 by root b plus root c 1 by root c plus root a and 1 by root a plus root b are in ap And see whether you reach a condition which is actually true. So you'll say let Let 1 by this b in ap Let this be in ap Which implies 1 by root c plus root a minus 1 by root b plus root c Is equal to 1 by root a plus root b minus 1 by root c plus root a So let me take the lcm over here of root c plus root a root b plus root c and on the numerator. I will have root b minus root a Similarly here also I can take the lcm and on the numerator. I will have root c minus root b Okay Now these two terms get cancelled And let me cross multiply So root b minus root a Multiplied to this which again is root b plus root a Will be root c minus root b times again root c plus root b Which is nothing but b minus a is equal to c minus b which is nothing but 2b equal to a plus c Which is actually true Because your a b c are in ap A b c are in ap and hence this assumption here is a correct assumption This is a correct assumption. That means these three terms are also in ap So we have been able to do the second problem also So this is also done Okay, now try the third one Try the third one again. So for the third one If a b c are in ap Okay, if a b c are in ap Uh, guys, will you all agree with me if I say that a by a b c b by a b c and c by a b c will also be in ap Remember the property where if you divide your all terms by a non-zero constant k the resultant sequence will also be in ap right Which implies One by b c one by a c and one by a b are also in ap Correct Now what I will do is I will multiply with a b plus b c plus c a throughout So I can say this three terms will also be in ap Okay Now again, I can write this as a b plus c a by b c plus one This I can write it as a b plus b c by a c plus one And this I can write it as b c plus c a by a b plus one Okay If I drop the factor of one if I drop this factor of one, can I say a b plus ac or a b plus c by b c And here I can say b a plus c by ac And here I can say c a plus b by a b. This will also be in ap This will also be in ap Okay, which clearly implies a one by b plus one by c b one by a plus one by c and c one by a Plus one by b are in ap and hence proved And hence proved. This is what I wanted to prove Okay These are very beautiful problems can come in your school exams as well. So be very very careful regarding these problems Remember to use properties wherever you can Wherever you can okay Now just one quick problem and then we'll move on to the geometric progression Just a small problem. I'll give you the problem is The question is the digits the digits of a positive integer Of a positive integer having three digits having three digits are in ap are in ap And their sum is 15 Their sum is 15 the number obtained the number obtained By reversing By reversing the digits Is 594 less than The original number Find the number Find the number So digits of a positive integer having three digits are in ap sum is 15 The number obtained by reversing the digits is 594 less than the original number Then find the number Yes, anybody any response 258 Among the reverse of it will be 852 852 is not less than 258 The question says is the number obtained by reversing the digits is 594 lesser than the original number 852 that's the correct answer absolutely So let's say the three digitted number Okay, is a minus d a a plus d Okay, the sum is 15. It implies 3a is 15, which means a is 5 Okay So the middle term will definitely be 5 now The original number will actually be a minus d into 100 a into 10 And a plus d let's say, okay This is the original number And when you reverse it it becomes a plus d into 100 a into 10 And a minus d So difference of this is 594 Difference of this is 594. So let us simplify this now So here I will get minus 2d times 100 And here I will get Plus 2d that's 594 Okay So I will end up getting minus 200 plus 2 In fact, you can drop the factor of 2 from everywhere minus 100d plus d Is equal to 297 So minus 99d is equal to 297. So d is going to be minus of 3 So d is going to be minus of 3 that means My number would be 5 mine Minus of 3 will become 8 over here And this will become 2 over here So 852 is the number 852 is the number So guys now we can take a break Just take a break till 11 25. So we'll have a break now Okay, we'll resume at We'll resume at 11 25 am sharp Just a five minutes break All right guys resuming back So now we are going to start with the geometric progression Now we are going to start with the gp That's the geometric progression All right, so what's the geometric progression geometric progression is a sequence where It's a sequence where The subsequent term differs from the preceding by a factor Right, so basically terms like a a r a r square a r cube, etc They will be in a gp Where you would realize that the ratio of the Terms preceding by succeeding is always a constant and that is actually called as the common ratio That's actually called a common ratio Okay, now again common ratio could be A positive it could be negative it could be greater than one it could be lesser than one But it cannot be zero r should not be zero in this case Okay, it could be imaginary also It could be imaginary also in that case your gp will become an imaginary gp Now if you continue this trend the n-th term of this particular sequence would be given as a r to the power n Minus one that is called the n-th term of an geometric progression Please note the similarity with an ap in an ap we had something a plus n minus 1d Now here what is happening the n minus 1 is raised on the power of r and a gets multiplied Okay That means if you start taking the log of terms in a gp so if a r a r square, etc, which are in gp If you take log of these terms that is log a log a plus r sorry a r log a r square, etc You would realize that this will become an ap Right, it's very important thing many a times it is asked in direct questions also So if certain number of terms are in gp then the log of those terms will be in ap So let us start with a question on this Let us start with a problem on the n-th term of a gp Question is if sin theta root 2 sin theta plus 1 and 6 sin theta plus 6 are in gp then The fifth term is Then the fifth term is option number a 81 option number b 81 root 2 option number c 162 Option number d 162 root 2 No, it's not in brackets. If I don't write a bracket. It's without bracket. It's 6 into sin theta plus 6 This is together and this is separate All right, so if we know three terms a b c are in gp then means b by a should be equal to c by b in other words b square is equal to ac Is this another one property which you all should note that in case of a gp In case of a gp the product of the terms The product of the terms equidistant Equidistant From the start and the end is constant And that is actually equal to the product of the first and the last That's equal to the product of the first and the last term product of the first and the last Okay, so b square which means 2 sin theta plus 1 the whole square is equal to sin theta times 162 is absolutely correct. So both shia and aryaman your answer is correct So I can always cancel out a factor of 3 from here and I can take sin theta plus 1 as common So I'll get sin theta plus 1 minus 3 sin theta Equal to zero that means sin theta could be minus 1 Or sin theta could be half So these are the two possibilities Okay, if sin theta is minus 1 what will happen this b term will become zero. So this is not possible So the only possibility is that your sin theta could be half So if sin theta is half first term is half second term would be two Root three by two And third term will become nine That means your common ratio is three root two By my bad my bad. I've written this as This becomes root two into three by two. It's three by root two. I'm sorry. Yeah And this will become nine which clearly implies Which clearly implies that your common ratio is going to be uh root two times three or three root two okay Common ratio is going to be this and first term is going to be half So fifth term is going to be a r to the power four, which is half into Three root two to the power of four, which is nothing but 81 into Two into two by two, which is nothing but 162 that's option number c is correct option number c is correct in this case Next question if abc are real numbers such that such that three times a square plus b square plus c square plus one is equal to twice a plus b plus c plus ab plus bc plus ca Then abc are in abc are in option number a ap only Option b gp only Option c gp and ap both Option d none of these Option d is none of these Any idea anybody? okay, so what I'll do here is I will write this term as two times a square plus b square plus c square And I will take this minus two ab plus bc plus ca term on the left hand side. Okay What remains is another set of a square plus b square plus c square c square term and you have minus 2a minus 2b minus 2c term and a plus 3 will remain. Okay, so let's say this is equal to 0. Now when you see this term carefully, it is actually nothing but a minus b square b minus c square and c minus a square, okay. And if you see this term carefully, you can actually generate a minus 1 square b minus 1 square and c minus 1 square, okay. Now square, each of these terms are perfect squares and they add up to give you 0. It means each one of them should be equal to 0 that means a is equal to b, b is equal to c, c is equal to a, a is equal to 1, b is equal to 1, c is equal to 1. So basically a, b, c all are equal to 1. That means these three numbers are in ap also and in gp also that means option number c becomes correct. So the next concept that we are going to talk about is sum of n terms of a geometric progression. So if you have a, ar, ar square and all the way till your nth term that is ar to the power n minus 1 and we have to find the sum of this. So what I will do is I will multiply this with r. So I will multiply the entire series with r. So a into r is ar. So I will write it one shifted to the right. So I will just shift this and write over here. So this r multiplies to a and I will shift it and write it over here, okay. Similarly, r multiplies to ar that is ar square and I will again write it over here. If I continue doing that, I would obtain this kind of a series. Let me subtract. Let me subtract. When you subtract a will remain over here. All these terms will get cancelled and I will get this term which means I can write s as a 1 minus r to the power n by 1 minus r. So please note that this can also be written as s ar to the power n minus 1 by r minus 1 as well, okay. Normally people say in fact in books it is written that use this when your mod r is less than 1 and use this when mod r is greater than 1. However, it doesn't make a difference. Both the formula are exactly the same, okay. Now something very interesting over here. What will happen when r is equal to 1? What will happen when your r is equal to 1? So we know some till n terms is this 1 r to the power n minus 1 minus r when mod r is less than 1. We also know that sum is ar to the power n minus 1 by r minus 1 when mod r is equal greater than 1. But what will happen when r is equal to 1? What will be the sum when r is equal to 1? So if you try to put it in any one of these formulas, you will get 0 by 0 form which is actually an indeterminate form. So in such cases what do we do? We take the limiting case. So we take the limiting case. So we take the limit r tending to 1 a 1 minus r to the power n by 1 minus r. Please try to recall that limit x tending to a x to the power n minus a to the power n by x minus a is n a to the power n minus 1. So similar situation is over here also. If you just pull out the a common, you will have limit r tending to 1, something like this r to the power 1 and 1 to the power n again r minus 1 which is actually n a to the power n minus 1 which is actually a n, okay. Which is obvious also because if you sum up a a a a a n number of times, you have to get a n. You have to get a n. Another way of rephrasing this result, another way of rephrasing this result, you can write this as s a to the power n minus a by r minus 1 which is actually a r to the power n minus 1 into r minus a by r minus 1 which is actually l r minus a by r minus 1. So if you know your last term, if you know your common ratio and if you know your first term, you can also write the sum like this. You can also write the sum like this. Now, if your mod r is actually less than 1, we call such kind of series, GP series as a convergent one, okay. Convergent means the terms will continuously start decreasing and the sum of that will limit to a fixed value. For example, something like this, this is actually a convergent series. This is actually a convergent series. In case of convergent series, we can actually end up finding the sum till infinity also. So if you have to find the sum of a convergent series till infinity, basically we can use this formula that is limit n tending to infinity of this. Remember, if mod r is less than 1, r to the power n will tend to 0 as n tends to infinity, okay. So this term will start becoming 0. So you will end up getting this as your sum till infinite term. So note that for a convergent geometric progression, the formula will slightly change and it becomes a by 1 minus r, okay. And if your mod r is greater than 1, the series, GP series will become a divergent one, okay. And in case of divergent one, you cannot find out the sum till infinity. It will either become infinity or it will become minus infinity, okay. So this is the formula for sum of infinite terms of a geometric progression. In case the geometric progression is a convergent one. So let us take some questions on this concept as well. First question is again if n, that is a set of natural numbers and be partitioned into sets s1 whose first term is 1, s2 which has 2, 3, s3 which has 4, 5, 6, 7, s4 which is 8, 9, 10, 11, 12, 13, 14, 15, then find the sum of numbers in s50. Find the sum of numbers in s50. Yes guys, any response? Yeah, express it in terms of 2 to the power something. RMM, no need to expand it and tell me. Just express it in terms of powers of 2. I am sure you would be getting powers of 2 in the answer. So 25 times, see first of all, you would realize that the number of terms, the number of terms in every set, let us say sk is nothing but 2 to the power k-1. So in s50, the number of terms would be 2 to the power 49 terms. And if you focus on the first term of every set, the first term of sk is actually 2 to the power k-1 again. Correct. So basically in s50, your terms would appear like this, 2 to the power 49, 2 to the power 49 plus 1, 2 to the power 49 plus 2, all the way till 2 to the power 49 terms like this. So of course, it's an AP whose first term is 2 to the power 49. Common difference is 1 and the number of terms itself is 2 to the power 49. So the sum will be nothing but 2 to the power 49 by 2 into 2a, that is 2 to the power 50 plus n minus 1 into common difference. That's going to be 2 to the power 48, 2 to the power 50 plus 2 to the power 49 minus 1. You can always take 2 to the power 49 common over here. So 2 to the power 48. If I take 2 to the power 49 common, I'll get 2 plus 1 minus 1. That's actually 2 to the power 48, 3 times 2 to the power 49 minus 1. So this will become your answer. This will become your answer for this question. That's correct. That's correct RMR. Yeah. Next question which comes very frequently in your school exams. Find some of the following series, 777, 777 and so on till n terms and something like 0.7, 0.77, 0.777 and so on till n terms. So let's do this question. All right. So I'll just do the first one so that you get an idea how to proceed in this case. In first one, please note that this is not a GP. So don't start applying the formula of a GP thinking that the common ratio is 11. 11 into 77 is not 777. Yeah, I know, I know. It's okay. Fine. No problem, Shreya. Yeah, you can send me a photo on the WhatsApp. So first of all, you take a 7 common which will be 1 plus 11 plus 111 and so on till n terms. Take a 9 outside like this. So it'll become 999 like this n terms. So I can write 9 as 10 minus 1, 99 as 100 minus 1, 999 as 1000 minus 1 and so on till n terms. So I can write this as 10 plus 100 plus 1000 and so on till n terms minus one collected n number of times which is this. Now, this is clearly a GP. This is clearly a GP whose first term is 10. Common ratio is also 10 and the number of terms are n. So I can write this like this, which is nothing but 7 by 9 10 times 10 to the power n minus 1 by 9 minus 1. If you take a 9 common out, it becomes 7 by 81, 10 to the power n plus 1 minus 9 n minus 10. This is your answer. In a similar way, we can proceed with the second problem as well. So I'll just give you a hint. I'll not solve it for you. I'll just give you a hint. In case of 7th problem, you can take 7 by 9 out. That'll actually leave you with 0.99, 0.999 and so on, which is actually 7 by 9, 1 minus 0.1, 1 minus 0.01, 1 minus 0.001 and so on, which is 7 by 9, n minus 0.1, 0.01, 0.001 etc. So you can treat this as a GP again with first term as this. Common ratio is 0.1 and 1 minus this is 0.9. So that gives you 7 by 9, n minus 1 by 9, 1 minus 0.1 to the power n. So this will be the answer for your second question. So this is the solution for your second question. Again, selection of terms in a GP. Selection of terms in a GP. So if you have to select three terms in a GP, preferably we'll select it as A by R, A, A, R, preferably. Again, I'm saying preferably it is not compulsorily. Preferably we'll select it like this. If you're asked to select four terms, preferably we'll select it like this. A by R, A, R, A, R, Q. If you're asked to select five terms, we'll select it as A by R square, A by R, A, A, R, A, R square. Six terms, A by R to the power 5, A by R cube, A by R, A R, A R cube, A R to the power 5. Like that. Okay. Especially where information about their product is known to us. Especially where the information about their product is known to us. Let's take a question on this. If the product of three terms, if the product of three terms in GP is 216 and the sum of their products in pairs and the sum of their products in pairs is 156, then find the sum of the three numbers, then find the sum of the three numbers, then find the sum of their three numbers. 26. Okay. Let's see what others have to say. RMR also says 26. Shalom. Most people are replying with 26. So again, if I have to take three numbers, preferably I would take it as A by R, A, A, R, and when I multiply it, I get 216, which means A cube is 216, which implies A is 6. Okay. And the product in pairs, which means A by R into A plus A into A R plus A by R into A R, that's equal to 156. That gives me A square by R and A square R, and this is A square, that is equal to 156. We already know what is our, what is our A? A is equal to 6. So A square is equal to 36. So this implies 36. R square plus R plus 1 by R is equal to 156. Okay. I can factor, I can cancel off a factor of 12. 12 will give me 3 and this will give me 13. So R square plus R plus 1 by R, that's equal to 13 by 3. Okay. So I'll get a quadratic in R over here. So 3R square plus 3R plus 3 is equal to 13R. So 3R square minus 10R plus 3 is equal to 0. So when we solve this 3R square minus 9R minus R plus 3 equal to 0. So 3R R minus 3 minus 1 R minus 3 equal to 0. So R could be 3 or R could be 1 by 3. So A is 6. If you take R as 3, you get the term before this as 2 and after this as 18. And if you take as R 1 third, this will become 18 and this will become 2. In any ways, the sum is going to be 2 plus 6 plus 18, which is equal to 26. That's absolutely correct. R M R was the first one to answer this. Sorry, Tiddi was the first one to answer this. Next question, there's an equilateral triangle. So I'll draw the figure over it. There's an equilateral triangle whose sites, whose one side is equal to 24 centimeter. Okay. Now what is done is we actually take the midpoint of the sites of this equilateral triangle. Okay. And connect them to form another triangle and keep on doing this. Keep on doing this. And this goes on and on for infinitely many times. Okay. So this goes indefinitely. The question is, the question is, find the sum of the parameters, find the sum of the parameters of all the triangles. Find the sum of the parameters of all the triangles. Okay. Tiddi says 144. R M R says 48 centimeter. How can that be possible R M R? Because 72 will be the parameter of the first triangle itself. So A is basically the side of the triangle, not the perimeter. Okay. So in this case, 72 will be the perimeter for the first one and exactly half of 72 will be the perimeter for the second one. And then 72 by four and so on till infinite terms. So basically if you have to find the sum of this, your answer will be A by 1 minus R. A is equal to 72. 1 by R is equal to 144 centimeters. So Tiddi is correct. So the answer for the first one is 144. Next question. If f is a function satisfying, if f is a function satisfying this functional equation, f x plus y is f of x into f of y, for all x comma y belonging to natural numbers, such that f of 1 is 3 and summation of f x for x equal to 1 till n is equal to 120. Find the value of n. Find the value of n. This is again your NCRT question. n is equal to 4 is what Tiddi says. Okay. Shea also backs him up. Okay. Gari also says the same thing. So guys, let's solve this. So if I have to find f of 2, which is actually f of 1 plus 1, I have to write it as f of 1 into f of 1, which means f of 2 is going to be 3 into 3, which is 9. Similarly, if I have to find f of 3, I have to write f of 2 into 1, which is f of 2 into f of 1. That's going to be 3 into 9, which is 27. So what I can see is that f 1 is 3, f 2 is 9, f 3 is 27, like that. It's following a GP. So this will be in a geometric progression. So some till n terms is 120. So some till n terms will be 3 to the power n minus 1 by 3 minus 1, which is 2, that's going to be 120. So this is going to be 80. So that implies 3 to the power n is 81. So n is equal to 4. Again, NCRT question. Next question is prove that 49, 4, 4, 8, 9, 4, 4, 8, 8, 9, etc., which is actually inserting, which is obtained by inserting, obtained by inserting 48 into the middle of the preceding number, into the middle of the preceding number is a square of an integer. So each of these terms are actually squares of some integers. For example, this is square of 7. This is the square of 67. This is the square of 667, etc. So in general, prove that that number will actually be a square of some number. If you're done, you can send me the proof on the WhatsApp also personally. If you see the trend, if I go to the nth term, if I go to the nth term, you would realize that your nth term will have n times 4. It will have n minus 1 times 8 and only 1, 9. Right? So I can write this as 9. Then we have 8 into 10, then 8 into 100, and so on till 8 into 10 to the power n minus 1, then we'll have 4 into 10 to the power n plus 4 into 10 to the power n plus 1, and so on till 4 into 10 to the power 2n minus 1. Yes or no? Is this correct? Yeah. Yeah. Now, this is nothing but a geometric progression. You can write this as a geometric progression, sum of a geometric progression whose first term is 80. Common ratio is 10, and number of terms are n minus 1. So I can write this like this. Similarly, this can be written as a geometric progression whose first term is 10 to the power n. Again, number of terms here is also n. Okay. So we can write it like this. Let's do one thing. Let's take a nine common in the denominator. So it will be 81. This will be 80 times 10 to the power n minus 1 minus 1 plus 4 into 10 to the power n into 10 to the power n minus 1. So if you open this up, it becomes 81 and minus 80 will get cancelled over here. So leave you with 180 into 10 to the power n minus 1. You can actually make it as 8 into 10 to the power n. You can make this as 8 into 10 to the power n. And here I will have 4 into 10 to the power 2n minus 4 into 10 to the power n by 9. Now, these two terms, I can club and write it as 1 plus 4 into 10 to the power n. And this will become 4 into 10 to the power 2n by 9. And this is clearly 2 into 10 to the power n plus 1 the whole square by 9. 9 is also 3 square. So we can write it as 2 into 10 to the power n plus 1 by 3 the whole square. Now, 2 into 10 to the power n plus 1 will always be divisible by 3. How do I show this? How do I show this? For all n belonging to let's say natural numbers, can we apply PMI principle of mathematical induction? So p1 will be what? p1 will be 21, which is clearly divisible by 3. So then we can say let pk be true for some k belonging to natural numbers and lesser than n. So 2 into 10 to the power k plus 1, let it be some 3 lambda. So 2 into 10 to the power k plus 1 plus 1 will be what? First of all, I can write this term as 10 into 2 into 10 to the power k plus 1. And this we can write it as 10. 2 into 10 to the power k from here I can write it as 3 lambda minus 1. So I can write it as 3 lambda minus 1. So if you open the brackets, you get minus 9, which is clearly divisible by 3. So which implies pk plus 1 is true. And hence by principle of mathematical induction and hence by principle of mathematical induction, Pn is true for all n belonging to natural number. So guys, we have been able to prove that this is going to be a perfect square of an integer. So that means it is going to be a perfect square of an integer and hence proved. Is that fine? Any questions with respect to this? Any question? Any concern? Please do ask right now. So next class when we meet, I think it will be a face-to-face class and we'll be doing harmonic progression as well as concept of means, AM, GM, HM, etc. And then a bit of introduction about AGP and then we're going to start with series. So I think one class more is required to finish off this topic. So guys, thank you so much for coming online today on a month. Over and out from Centrum Academy. Have a good day.