 Hello all being from Cinema Academy. Welcome to the YouTube Live session on sequence and series. So those who have joined in the session, I would request you to type in your names in the chat box. So this is going to be another problem solving session on sequence and series chapter. Hello, good evening, Shruti, Sai, Atmesh, Vishist. Welcome to this live session. So let us begin with the first problem of the day. So here we have the very first problem coming up for you. So the question says, let A1, A2, A3, et cetera, be an arithmetic progression such that 1 by A1, A2 plus 1 by A2, A3, et cetera, till 1 by A4000. A4000, 1 is equal to 10. And it is also given that A2 plus A400 is 50. Then you have to find the mod of A1 minus A4000 and 1. I think there's slight typo error over here. This has to be A4001 over here. So please make that change. So it's finally, this term is actually A2 plus A4001 is equal to 50. So this is A4001. Make this correction over here. Welcome, Ramcharan, Vaishnavi, Sanjana, Shreyas, Swayatya, good afternoon to you all. So again, we'll keep the rules simple, just like the last time. I should at least get four responses so that I can start solving the session. And please feel free to type in your response in the chat box. OK, so we have the first response coming from Ramcharan. Noted, Ramcharan. So let's wait for three more people to respond. Just a small change again. I think there is slight error. This has to be 4,000, not 4,001. Else it'll become slightly difficult for you to solve the problem. So there was just a typing error. So make that A2 plus A4000 is equal to 50. So Ramcharan, do you want to stick to that answer or would you like to change it? So those who have joined in the session request you to type in your names in the chat box. Hello, welcome, Sondarya to the session. All right, so let's start the discussion. There's already six minutes gone and I've just got one response. So let's start solving the session. Now here, when I write 1 by A1, A2 plus 1 by A2, A3, et cetera, till 1 by A4000, A4000, 1. Let's say it is equal to 10 given to us. Now what I'll do now is I will multiply both sides with D. So I'll multiply on both sides with D. So when I do that, so I get D by A1, A2 plus D by A2, A3, and so on till D by A4000, A4000, and 1. Now this D over here, I will write this D as A2 minus A1. This D over here, I would write it as A3 minus A2. And if I keep doing this till the last D, which I can write it as A4000, 1 minus A4000, I know that I can use here my method of difference. So I can write this as 1 by A1 minus 1 by A2 plus 1 by A2 minus 1 by A3. And this goes on till 1 by A4000 minus 1 by A4000, and 1 is equal to 10 D. Now terms start getting cancelled like this. And ultimately what is left off is 1 by A1 minus 1 by A4000, and 1, that is going to be 10 D. Is that fine? So I can actually write it as A4000, 1 minus A1 by A1, A4000, 1 is equal to 10 D. Now A4000 and 1 can be written as A1 plus 4,000 D. So this numerator term, I'm just changing it to 4,000 D. This will become 4,000 D. So D and D will get cancelled off. And I will get A1, A4000, and 1 as 400. Is that fine? OK. Now once we have got that, we can make use of this information, which is as good as saying A1 plus A4000, 1 is equal to 50. Please remember in an arithmetic progression, the property is the terms equidistant from the beginning and the end, the sum of those two terms are constant. And that sum is equal to sum of the first and the last. Now what is the requirement of the question? The requirement of the question is mod A1 minus A4000 and 1. So what I'll do is I'll make some place for myself over here. I'll just erase a bit of it. Let's say I erase this part. Now we know that if I have to find this, basically I have to do under root of A1 plus A4000, 1 whole square, minus 4A1, A4000, and 1. So that is going to be under root of 50 square minus 4 times 400. That's going to be under root of 900, which is going to be 30. So option number B here becomes your right answer. Let's see who was the first one. I think the first one to answer this correctly was Ramcharan and Saimir. So two right answers I got for this particular question. Is that fine, guys? Any question, any concern? Please do type in in the chat box. Easy question to start with. All right. Let's move on to the second question. So here we have a question that there is an AP which consists of even number of terms having middle term equal to, in fact, middle terms. So they're two middle terms and they're equal to 1 and 7 respectively. If n is the maximum value which satisfies this equation, A1, A2n plus 713 is greater than equal to 0, then the value of the first term of the series, then the value of the first term of the sequence or series, is which of the following options? Yes, guys, it's very important to be accurate. So you may take a minute or two extra, but try to get a correct answer as far as possible. I tried to identify the clues in the question. The clues in the question is we have to find out common difference. We have to find out the value of n. OK. And probably that would lead us to the value of the first term. OK, so I've got a response from Lahitya. OK, Rohan also. Guys, this chapter is very easy and scoring. And every year, at least one question comes from this chapter in JEE main exam. OK, so I've got three responses so far. One more I need, so that I can start solving this question. Wishes, Shriyas, Sanjana, Ramcharan, Shruti, Sondhria, I'm waiting for your response. Atmesh, OK. Atmesh also is, so all of you have answered with option D. OK, let's see what happens. So guys, if you know if the AP consists of even number of terms, your middle term would be your nth term and your n plus 1th term, right? So nth term and n plus 1th term would be your middle terms. And they are given to us as 1 and 7, respectively. So these two will be your middle terms, undoubtedly. Right? OK. Now, I also know for the fact that the sum of the two middle terms is as good as the sum of the first and the last term. Remember the properties which we had discussed in the previous question as well? That is, in an arithmetic progression, the sum of terms equidistant from the beginning and the end is a constant and that is equal to the sum of the first and the last term. Right? OK. Apart from that, we also know that the common difference has to be 6, isn't it? Because the common difference has to be the n plus 1th term minus the nth term. That's going to be 7 minus 1, which is equal to 6. Is that fine? All right. So using this information that t1 plus t2n, OK, which is as good as tn plus tn plus 1, that's going to be 8, 1 plus 7, which is going to be 8. So let the, let a be the first term. Let a be the first term, OK? In other words, what I've been given to us is a plus n minus 1d and a plus nd adds up to give you 8, right? Right? So if you collect the a's, it'll be 2a plus 2n minus 1d is equal to 8, right? So far, so good. OK. So now 2a remains 2a, but this I can write it as 2n minus 1 into 6, correct? And that is equal to 8, right? We know the value of d, right? So d6, correct? So from here, I can say a plus 2n minus 1 into 3 is equal to 4. That means a plus 6n is equal to 7. Let this be one of my equations, OK? Let me call it as 1. Now, let us make use of the other information that is given to us. That is a1 into a2n is plus 713 is greater than equal to 0. Now, we all know that 4a t1 into a t2n would always be following this identity, which is t1 plus t2n whole square minus, you can say, t1 minus t2n whole square, right? And t1 plus t2n is as good as tn plus tn plus 1. That we have already decided, and that is going to be 8 again. So this is going to be 64, right? OK. And a t1 minus a2n square is going to be, if I'm not wrong, a2n can be written as a t1 plus 2n minus 1 into 6. So this could be written as t2n minus t1 equal to 2n minus 1, 6, the whole square. So if you square this, you are going to square this as well. OK? So what I'm going to do is I'm going to replace it over here and write this term as 36. In fact, minus sign will be here in between. Minus 36 2n minus 1, the whole square. Is that fine? OK. Now ultimately, what is given to us? Let's use that, divide by 4 throughout. So we'll get 16 minus 9 2n minus 1, the whole square, plus 713 is greater than or equal to 0. Correct? Let's solve this inequality. So I can write this as 729 minus 9 2n minus 1 whole square is greater than or equal to 0. Drop the factor of 9 from both the sides, right? So drop the factor of 9, so it will become this. Correct? Now we can also write it as 2n minus 1 whole square minus 9 square is less than or equal to 0. So it can be factorized as 2n minus 1 square minus 9 and 2n minus 1 plus 9 less than or equal to 0, right? Now let me make some space for myself. I'll just erase the top part of this because we have already made use of this information. And now I'm going to solve this inequality 2n minus 10 and 2n plus 8 less than or equal to 0. So drop the factor of 2 as well. So n minus 5, n plus 4 is less than or equal to 0. So this implies n lies between minus 4 and 5. So the maximum value that satisfies this relation is n is equal to 5. The moment I get n equal to 5, I can make use of the first equation. So from 1, I can say a plus 6 into 5 is equal to 7. So a plus 30 is going to be 7. So this implies my a will be minus of 23. That's option number D, which is going to be correct. And let's see who was the first person to answer this correctly. The first person to answer this correctly was law hitya, awesome law hitya, very good. I would say this was not an easy problem because it involved a lot of manipulations here and there. So guys, let me tell you, Jay is known for framing difficult questions on easy topics because they know people are going to take it very lightly. So they are trying to take advantage of every loophole that is possible. Is that fine? Any question, any concern? Please do ask me in the chat box. Chalo, great. So we'll move on to the next question, which is the third question for the day. So here we are with the third question. If the sum of the first 100 terms of an AP is minus 1 and the sum of even terms lying in the first 100 terms is 1, then which of the following option is not true? Is not true? So read the question very, very carefully. So which of the following option is not true is what we need to answer. So guys, I understand your boards are going to start very soon, but do not discontinue writing mock tests, which will be sending you also. And of course, keep writing mock tests of board level as well. That is something which I am also going to post on the group. Now guys, if you're giving the option for this question, please type the question number as well, because then it will not be confusing for me that whether you have answered the previous one or the new one. So let's say if you think the answer for this is C, then type 3, bracket C. So we need to comment on the common difference, the first term and the 100th term of this arithmetic progression. OK, so wishes has given a response. How about others? Rohan, Laitya, Sanjana, Shriyas, Shruti, Sondarya, Ramcharan. I need three people to respond, at least, so that I can start solving this problem. OK. One more required. All right, so let's begin with the solution of this problem. So first thing that has given to me is the sum of the first 100 terms of the AP is minus 1. So I can say some of the first 100 terms will be equal to 100 by 2. And let's say our first term is x1 and the 100 term is 100. OK. So that's going to be 100 by 2. x1 plus x100 is equal to this is going to be minus of 1. Correct. Which clearly implies x1 plus x101, sorry, x1 plus x100 is going to be minus 1 by 50. OK, let me call this as the first equation. So since I have to find first term and 100 term, it's better to keep things pretty simple. So whatever is the requirement, just keep your expressions in those terms only. Next, what is given is a2 plus a4 and all the way till a100. That means some of all even terms lying in the first 100. That's a2 plus a4 all the way till x100. And that is given to us as 1. By the way, I can see that there will be 50 terms like that. So 50 by 2. x2 will be x1 plus d. And again, x100, that's going to be 1. So this gives me x1 plus x100 plus d is equal to 1 by 25. Let me call it as the second equation. Now, undoubtedly, I already have x1 plus x100 as minus 1 by 50. And I have x1 plus x100 plus d as 1 by 25. So this clearly implies if I subtract, if I do 1 minus 2, it will lead to d that is going to be 1 by 25 plus 1 by 50. And that is nothing but that's going to be 3 by 50. So the common difference is going to be 3 by 50. So option number A is correct. So it cannot be my right option. Next is I have to focus on the first term. Now, if I know x1 plus x100, that means I know x1 plus x1 plus 99 d, which is actually this. And that's minus 1 by 50. Correct. So 2x1 is equal to minus 1 minus 297 by 50. So that's going to be minus 149 by 50. So first term of the sequence is minus 149 by 50. This is also a true option. So it cannot be my right option. So 100 term will be very simple. Again, we know that this is your first term. So add 99 into d with that. So that's going to be again 297 minus 149 by 50. Let me write it properly. Minus 149 by 50. So that is going to be around 148 by 50. That's going to be 74 upon 25. That means option C also is a correct choice. I mean, the information is correct. So it cannot be my right option as well. So what remains is option D. So none of these options are actually wrong. All of these options, all of these informations cited to me are correct. So correct option, which is not the true option, is the option number D. So the first one to answer this question correctly was, I think, Vishishth. Vishishth saw was the first person to answer this correctly. Well done, Vishishth. Very good. Guys, I don't think so. This was a difficult question. I was expecting a response well within three minutes for this question. Anyway, let me move on to the fourth question for the day. So here we have the question given at the sequence of numbers x1, x2, till x2005, which satisfies this given condition, x1 by x1 plus 1 is equal to x2 by x2 plus 3 is equal to x3 by x3 plus 5, and so on till x2005 by x2005 plus 4009. Find the nature of the sequence. Find the nature of the sequence. All right, so got a response from Vaishnavi and Vishishth so far. All right, shares. Saim here also join in with the same option. Okay, guys, so let's discuss this now. I've got a response from four people. So what I'll do is I'll consider this entire series to be equal to a fixed value. Let's say that value is lambda. Okay, so let's say this is equal to da-da-da-da-da-da, and this final expression, they all are equal to lambda. Okay? So now let us work with the first expression equated to lambda. So let's say this is lambda. So I can say x1 is lambda, x1 plus lambda. So I can take x1 common. So 1 minus lambda is equal to lambda. So x1 is equal to lambda by 1 minus lambda, right? If you look at the second expression, that is x2 by x2 plus 3 is equal to lambda. Okay, so let's see what do we get from here? So x2 is equal to lambda x2 plus 3 lambda. That implies x2 1 minus lambda is 3 lambda. So x2 is equal to 3 lambda by 1 minus lambda. Okay, let me compare the third one as well, because from two I cannot make any conclusions because it looks like a GP as of now. Okay? With a common ratio of 3. But let's see what do we have when I compare the third expression with lambda? So again, x3 could be written as 5 lambda by 1 minus lambda. Oh, thank God. I did not make a judgment that is a GP because I could see the terms to be lambda 1 minus lambda, 3 lambda by 1 minus lambda, 5 lambda by 1 minus lambda, et cetera, which is clearly an arithmetic progression, right? With the first term as lambda 1 minus lambda and common difference as 2 lambda 1 minus lambda, okay? So it has to be an AP. So option number A is correct. And the first person to answer this correctly was Vaishnavi. Well done, Vaishnavi, very good. Okay, again, a simple question, but yes. Guys, you need to be fast, right? Can you move on to the next one? So let's take up the next one, which is the fifth question for the day. So this question is A, B, C are positive integers forming an increasing GP. And B minus A is a perfect cube, okay? So B minus A is a perfect cube. And log of A to the base six, log of B to the base six, log of C to the base six is equal to six. Then find the value of A plus B plus C, okay? Vicious, again, has given the response first. Need three more responses, okay? So Rohan also agrees with what Vicious has to say. Well, good. Others, Suresh, Ramcharan, Sanjana, Tye, Sondarya, Shruti, what's up guys? Yes, guys, needs two more response. All right, so first thing first, we'll make use of this property that we have over here. I can say that log of A, B, C to the base of six is going to be six, which clearly implies your A, B, C is going to be six to the power six, right? Okay, and these terms are in GP, right? It's better to take the terms in GP as B, B by R and B R, because now we have information about the products. So let's say this is your A and this is your C. So when you multiply them all, this gives you six to the power six, which actually means B cube is six to the power six, which means B is six to the power two, which is 36, right? So your terms are 36 by R, 36 and 36 R. Now, next information that we have is B minus A is a perfect cube. So 36 minus 36 by R is a perfect cube. So it's a perfect cube, okay? So if it is a perfect cube, one possibility is R can be one because zero cube is a perfect cube, right? Now, since we are all dealing with integers, okay? So the next that we can have, let's say choose RS2. So when you choose RS2, you get 18, which is not a perfect cube. If you choose RS3, I get 24, that is also not a perfect cube. But if I choose RS4, I get 39 minus nine, 36 minus nine, which is 27, which is a perfect cube. So R has to be four, right? So make sure that you use this information. It's an increasing GP, and you have positive integers as your terms of the GP, right? So only possibility is RS4. Now, if RS4, it's obvious that we know all these terms, that will be 9, 36, and 36 into four is 144. So when you add up, it's going to be 144 plus 45, that's 189 is going to be your answer, which is option number D is correct. Option number D is correct. I'm really surprised to see only two people answering this. Guys, I'm sure your keyboard at home is working, or if you're watching it over your phone, I think you can type in, right? This was a super simple question, okay? I was expecting everyone to give the response here. Anyways, I will move on to the sixth question for the day. So please read this question. The first three terms of a geometric sequence are X, Y, Z, and these have sums equal to 42. If the middle term Y is multiplied by five by four, then these three numbers, X, five by four, and Z now form an AP. Find the largest possible value of X. Find the largest possible value of X. All right, so Sanjana, Lohitya, Vishis, almost everybody replying at the same time, six C is what they say. Okay, so let's look into this now. So the first term of a geometric progression, first three terms are given as this, and they have the sum equal to 42, right? So I can consider the terms to be Y, Y by R, and YR, and the sum is given to us as 42. If the middle term is multiplied with five by four, then these numbers, then these numbers, they form a arithmetic progression. So twice of this is equal to X plus Z, okay, correct? So if you drop the factor of Y from both the sides, I can say five by two is equal to R plus one by R, which clearly implies R can take two values, one is two, other is half, correct? Now from here I can say Y one by R plus one plus R is equal to 42. So if I take the value of R as, or this as five by two, so this becomes seven by two is equal to 42, which means Y is equal to 12, correct? So Y is equal to 12. So the possible value of X could be, the possible value of X could be 12 by two or 12 by half. Out of this, I think the maximum is 24. So very simple question, option number C is correct, and the first one to answer this correctly was Sanjana, well done. So guys, moving on to the seventh question, and infinite GP has second term as X and its sum is four, then X belongs to, all right? So wishes have already given a response. All right, so guys, I'm getting all types of answers. I think only B option has not been said by anybody. Apart from that, all the answer options have been given. Anyway, so let's look into this question. Okay, so let's say the second term is X. Okay, so if second term is X, then this clearly implies the first term would be X by R, assuming R is the common ratio, right? So what is the sum of an infinite GP? We know some of the infinite GP is the first term divided by one minus common ratio. So that's going to be X by R divided by one minus R, that's going to be equal to four. So undoubtedly my X is going to be four R, one minus R. Correct? Right? Now, if I look at this term, R one minus R, okay? It's going to be basically a, let's say I call this as a function of R, okay? So if I draw the graph of this function of R, I have all the tools with me to draw the graph, so why not to use it? So let's say I have to draw this graph. So it is clearly going to be a parabola which is opening downwards like this, correct? Passing through zero and one. So this is zero and this is one, correct? Now remember, when you're talking about a geometric progression, your mod R has to be less than one. That means your R cannot exceed this interval. Is that fine? Okay. So you have only to see the maximum and the minimum value of the function in this interval, that is minus one to one. Okay, so this is your one, minus one to one. Is that fine? Okay. So what are the minimum value? Minimum value will be the value which it attains at minus of one. This is the minimum value I can say, okay? Maximum value is the value that it will attain at half. Isn't it? Maximum value will be the value that this fellow will attain at half, right? So your FR minimum would be minus one, one plus one, which is actually minus of two. And FR maximum would be a half, one minus half, right? That's going to be one by four if I'm not mistaken. Okay? Now my X is going to be four times, the minimum value of X is going to be four times the minimum value of this, which is minus eight. And the X max value is going to be four times this, which is one. So X belongs to the interval minus eight to one, okay? So option number C is going to be the right answer for this. And I think the first person to answer this correctly again was Vishishth. Easy question, but again framed in a very tricky manner. Great. So let's move on to the next question now, which is question number eight for the day. So the question says in a GP, the ratio of the first 11 terms, I think they, they mean to say last 11 terms. This is not even terms, this is 11 terms. So yeah, is the ratio of the first 11 terms to the last 11 terms is one by eight. And the ratio of the sum of all the terms without the first nine to the sum of all the terms without the last nine is equal to two. Then the number of terms in the GP because minus eight actually comes exactly when R is equal to minus one, right? But in the interval minus one is not included. R cannot be minus one. Okay. So in order to be a convergent geometric progression, R cannot take a value of minus one. So minus one was the one corresponding to minus eight. That's why minus one was excluded. Is that fine, Vaishnavi? Okay. So I've got a response from Rohan, Rohan wishes. They are the early risers. Others are still sleeping guys, baka fast. I need two more response. I know there's too much of information dumped on you, but you must learn how to navigate with so much of information. Okay. So Sai has a different answer. I is one more person, one more person and I'll be ready to start this problem solving. All right. So let's start with this problem. So what we have been given is that the sum of the first 11 terms is, to the sum of the last 11 term is one by eight. So we know some of first 11 terms is going to be A. Let's say I assume R to be one minus R to the power 11. R is the common ratio. And let's say some of the last 11 terms. Guys remember when you say last 11 terms, you can actually start with the first term as A R to the power N minus 11. One minus R to the power 11 by one minus R. So if you take the ratio of these two, it becomes one by eight. So the ratio of these two will be simply A by A R to the power N minus 11 and this is one by eight. Okay. So the first equation that we get here is R to the power N minus 11 is eight. Now let's move on. Let's see what the other statements have to provide us with. Then the ratio of the sum of all terms without the first nine terms to the sum of all terms without the last nine. So if you take the ratio of the terms without the first nine term, it is going to be, so N minus N terms will remain and this will be your sum of all terms without the first nine term, right? So let's say S dash nine, okay? And the sum of all the terms without the last nine will be this, correct? And this ratio of these two terms is given to us as two. So if I divide these two, that means I divide, let's say I call this as one and I call this S two. If I divide one by two, I'll get R to the power nine as two, correct? Now if I look at this question, if I look at this expression, R to the power N minus 11 is eight and R to the power nine is two, correct? Which means R to the power nine cube is going to be eight, which I can compare with R to the power nine minus 11. So R to the power 27 is R to the power N minus 11. Now I can easily compare the powers. So N minus 11 is going to be 27. So N is going to be 38. So altogether, the number of terms in this sequence is going to be 38 terms. And again, let us see who was the first one to answer this. So the first one to answer this was a Rohan, right? That was fast. Good, good one, well done. Is this fine now? So guys who made a mistake, please analyze where you went wrong. I think it was just in the matter of writing the terms correctly, great. So we'll now move on to the next problem. That's problem number nine. So we have been given the arithmetic mean of two positive number is six and their geometric mean is G. Harmonic mean is H and satisfies the relation G square plus three H is equal to 48. Then the product of the two positive number is. I think you cannot get a simpler question than this. This is super simple. I'm expecting answers to come within a minute. So less than a minute, you should be able to answer this. And here it is, yeah, so I got the first response from Sai. All right, yeah. So everybody's agreeing with B. So guys, let the two positive numbers be A and B, right? That's the hint for you. Okay, so Lohitya, what does that yes mean? Do you agree with what they have to say? Okay, nine B is what people are saying. Okay, so let us see whether Janta is correct or not. Guys, we know that if A is the arithmetic mean, G is the geometric mean, and H is the harmonic mean of two positive numbers, then remember that themselves will be in GP. Okay, that means AM, GM, and HM of two positive numbers themselves are in GP. Isn't it? Correct? Now, so G square will be arithmetic mean. That's going to be six, okay? So six is our arithmetic mean. And you can also say that A plus B by two is known to us, which is equal to six. And harmonic mean is two AB by A plus B, okay? So we know A plus B from this equation, that means A plus B will be equal to 12. And from here, I can say G square is going to be 12 AB by 12, right? So G square is going to be 12, undoubtedly. What am I saying? One second. That's going to be AB. Yeah. Now, three H, three H is going to be two AB by A plus B into three. That's again going to be six AB by 12. That's going to be AB by two. So what is given to us in the problem is AB plus AB by two is equal to 48, which means three AB by two is equal to 48, which means AB is going to be 32. So again, option number B here becomes correct, okay? And the first one to answer this correctly was Psi. Okay, great, guys. So I think this problem was very, very easy problem. So it doesn't need any explanation, any clarification. So let's move on to the next question. Moving on to the 10th question. So XYZB three terms in GP says that four is the AM between X and Y and nine is the HM between Y and Z. Then find the value of Y, okay? So I'm getting, I've got two responses. Others, guys, I need two more response and then I'll solve this problem. All right. Yeah, so let's solve this problem. So we have been given that XYZ are three numbers in GP. Okay, so I can say, let the terms be Y, Y by R and YR. Okay, so let X be Y by R and Z be YR. So that four is the AM between X and Y. So arithmetic mean of X and Y will be Y by R plus Y by two. So that's going to be eight for you, okay? And nine is the harmonic mean between Y and Z. Okay, so harmonic mean between Y and Z will be two AB by A plus B and that is given to you as nine, which means two YR by one plus R is going to be nine. Now from these two equations, first and second, I need to find the value of Y, okay? I can easily do that. I can actually replace, I can actually write this as Y plus YR is equal to eight R. And from here, I can say Y could be written, in fact, R could be written as Y is equal to R eight minus Y, okay? So R is equal to Y by eight minus Y. So putting it over here, I get two Y into Y by eight minus Y is equal to nine. One plus R, one plus R will be eight by eight minus Y, which clearly means I can say Y square is equal to 36, which means Y is equal to plus minus six. So here six could be a possible answer, okay? And remember, we deal with positive numbers, okay? And Y cannot be eight, Y cannot be eight. So option B is a possibility for a right answer. Is that fine? So all right, guys, before moving on to the next question, I would take a break over here, okay? Okay, so all of you, please take a break. We'll be resuming at 5.30 p.m. All right, guys, welcome back. Moving to the 11th question for the day, again, a simple one. If the harmonic mean of half one by two square, one by three cube, sorry, one by two cube, all the way till one by two to the part 10 is lambda by two to the part 10 minus one, then the lambda value is, okay? Sai says option A, all right? So Vishis also thinks the same, Sanjana also, fine. I need the response of one more person, okay? So Rohan gives a different answer, he thinks 11 D. All right, guys, so we know that, we all know that if we have terms like A1, A2, A3, et cetera till AN, okay? Then the harmonic mean is given by the total number of terms by reciprocal of each of these terms summation, right? So this is the harmonic mean formula of N terms, right? So in the present question, I can say my harmonic mean would be equal to, total number of terms are 10, okay? Reciprocal of these terms would themselves be two, two square, two cube, all the way till two to the part 10, which in itself is a geometric progression, right? So in geometric progression, we can say the sum is A common ratio to the power of N minus one by two minus one, isn't it? Which means it becomes five by two to the part 10 minus one and which we have to compare with lambda two to the part 10 minus one. So your lambda value here is going to be five. The lambda value here is going to be five. So none of you gave the right answer. Option B is correct. My God, this is a surprise. Cannot believe this. This was a sitter easy question. Oh God, this was a shock. Anyways, let's move on to question number 12. An aeroplane flies around squares whose all sides are of length, 100 miles. If the aeroplane covers at a speed of 100 mile, the first side, 200 miles per hour, the second, 300 miles per hour, the third and 400 miles per hour, the fourth, find the average speed of the aeroplane around the square. Yeah, it's a very easy question guys, very easy question. All right, so almost everybody has given the response and I think all of them, all of you are correct. So the average speed of it will be four by, it's actually nothing but the harmonic mean of the speeds that's going to be 400 by, you can say three by two plus seven by 12. And that's going to be 400 into 12 by 618, 25. Okay, that's going to be 16, so 16 into 12, that's going to be 192 miles per hour. So option C is correct. Okay, easy one again. Next, question number 13. So if the positive integers are written in a triangular array as shown, then the row in which the number 2010 will be. Okay, notice on the left, let's see what others have to say. Okay, Rohan say something else. Okay, Sai, I need one more response. Yes, Shruti, Ramcharan, Shreyas, Atmesh. Okay guys, so let's discuss this question. See, I'll assume that let the K-throw, let the K-throw contain 2010. If you look at the first terms of every row, right? They're like this, two, four, one, two, four, seven. Probably the next one will be 11, et cetera. Now, if I ask you, if such a pattern continues, what would be the first term, what would be the first term of the K-throw? How will you find this out? Now, here I had told you method of difference, right? Method of difference was a method which we had discussed in the class that if the difference of the terms, right? If it comes in an AP, right? Then we use something called the method of difference or differencing method, where we assume that the nth term is actually a quadratic in N, right? If the first set of terms themselves are in AP, we treat the nth term as a linear term in N, right? If the second difference is AP, it's a quadratic in N. If the second, third set of difference is basically in AP, then it will be a cubic in N and so on and so forth. Now, how do I get this value of A, B and C? Very simple. Put N as one and equate it to one. Put N as two, equate it to two and put N as three and equate it to four, correct? Now, solve these three simultaneously. It's very easy to solve them. Just take the difference. So, three A plus B is going to be one and five A plus B is going to be two. Now, again, take the difference. So, two A is going to be one. So, A is going to be half, okay? So, if A is half, B is going to be minus half and C is going to be one. That means, undoubtedly, the nth term of this particular sequence will be N square by two minus N by two plus one, okay? Or you can say N square minus N plus two by two, okay? So, if I say Kth term, the first term of the Kth series it will be TK. So, I can just replace my K throughout and throughout with K. So, it will become K square minus K plus two by two. Similarly, the first term of the K plus 1th row will be K plus one square minus K plus one plus two by two. Correct? Which if you open, you'll get K square plus K plus two by two. Now, if 2001 has to lie in the Kth row, do you agree with me that 2010, yes or no? Is that fine? Now, let us solve these two in equations simultaneously and see what is the value of, or what is the integral value of K that we get? So, K should be an natural number in this case because we are representing the row position with respect to that. So, so far it is clear, right? Can I make some space by erasing these parts? Let's erase this part, okay? So, let me erase this entire thing. So, we'll take the first situation. So, K square minus K, and I think if you multiply it to it becomes 4,020 and take it to the left side, so it would become 4,018 and it should be less than equal to zero. Similarly, here I would get K square plus K minus 4,018 would be greater than zero. Correct? Here, if you want, you can complete the square, K minus half the whole square and this would be less than equal to 4,018 plus one by four, okay? So, that would be approximately K minus half whole square would be less than equal to 1,070, 16,073 by four, okay? Similarly, here I can say K plus half the whole square is greater than 16,073 by four, okay? So, if I try to take the under root, guys, any rough idea how would, what would be the value of K minus half when this is under root of this? Any approximate idea that you have, okay? Any idea? What would be the approximate value of under root of 16,073 by four? That means this. So, start making pairs. I think the process of under root has to be understood over here, okay? One, okay? It's near about 165, 22, yeah, 22, so it's 44. So, 16,73, so that's going to be 24 now. Roughly six, seven, seven would work. I think 247 into seven will be 9, 4, 28 and 2, 3, 17, oh. So, roughly six, okay? So, it'll be more than 126 point something, right? So, it'll be 63 point something, correct? So, K minus half should be 63 point something and here K plus half should be greater than 63 point something, correct? So, basically I can say that K is sandwiched between, K is somewhere around 63, right? So, 61 is not possible. So, 63 is the exact, I mean to be very precise, 63 is the value that 2010 should belong to. So, the row number should be 63 in this case. So, guys, this question was not tricky, I would say, but yes, it was a bit of calculation intensive can waste time, just like the previous GE, many of you experienced that you wasted a lot of time doing calculations, okay? So, you must learn how to take approximations while doing it. You don't have to exactly sit and find two decimal or three decimal places. Just work with approximations. Just see which of the option is best suited to that particular interval that you are getting, okay? So, let's see who was the first one to answer this. I think the first one to answer this was Rohan. Oh, that was pretty fast. He answered this question pretty fast. Well done, Rohan. Very good. Let's move on to question number, question number 14. So, we have a summation of a summation of a summation of one. Just be careful about the inequality sign over here. I think this was I. I think they have written K, my mistake. Sorry, I is in the inner loop. I'm sorry. This is I, okay? So, I'll rewrite this summation K equal to one till 10, J equal to one till 10. It doesn't make a difference though, but of one. When I is less than J, J is less than K. 14 D, that's what Vaishnavi and Vishis say. Okay, just to tell you, 14 D is not the right answer. You guys think very, very carefully. Don't be in a hurry to answer this. This is a symmetric summation on I, J and K. So, make use of that. Anyone? I think if Nishal would have been in the session, he would have quickly written a code, especially in Python, it's very easy to write a code for such conditional loops. You're saying I should be in the outer loop. Why? Any specific reason why you think it should be in the outer loop? Moreover, will it make a difference? Will it make a difference if you change I with J, J with K? Okay guys, let's try to understand this. I'll first give you a simpler question than this itself. If I just give you this question, summation of, summation of, summation of, what will be the answer for this? What will be the answer for this first of all? So all of you please type in in your chat box. What will be the answer for this? There's no restriction. I is going from one to 10, again J is going from one to 10, and again K is going from one to 10. All right, the answer will be 1000. No doubt about it. Now, next question that I would like to ask you, what if I put this restriction? What if I say that my I cannot be equal to J and J cannot be equal to K? That means all the three I, J, K cannot be the same. Then how would the answer change? How would this answer change? So what I will do is, I will first take those conditions where there is no restriction, which is 1000, correct? And subtract those cases, subtract those cases where any two of them are equal. Let's say, let's say I take a case where K is equal to J. Right? So K is equal to J is the case which I take over here, okay? Similarly, I will also have a case where I can have I is equal to J, and I can also have a case where K is equal to I, correct? Now tell me what will happen in such summations where you are assuming two of them are same? That means, let's say K and J are same. That means these two are same. So is it not akin to saying that if two of them are same, it will be transformed to as if you are just summing up one from I equal to one till 10 and J equal to 10, right? Because K and J are same, isn't it? So can I say there will be three such cases like this, one for this, two for this, and three for this, and each of these three cases, your answer will be 100, isn't it? So I have to subtract 300 from here, right? But while you are doing that, you may also end up getting, but remember we have to also subtract that case where all of them are same, right? But that is already subtracted three times over here. So in this, you have already subtracted three times that case where all of them are same. So you have to add two such cases where all of them are same, right? It's like, you know, you are dealing with sets, right? So in set, you have already subtracted A union, B union, C union, A. But while you're doing that, you have already subtracted three times the intersection of all of them, isn't it? But that ideally had to be subtracted only once, isn't it? So you're adding two times that, and this 10 is because it's like summation of one from one to 10, because all of them are equal, right? I and J and K all are equal here. So it's just 10, correct? So when I is not equal to J is not equal to K, your answer comes out to be 1000 minus 300 plus 20, which is 720. Now, when you say I is less than J, J is less than K, it actually is one of the conditions where I, J and K are not equal, isn't it? However, there can be many such conditions where I, J and K are not equal, okay? For example, you can have I less than J less than K, you can have I less than K less than J, you can have J less than I less than K, you have J less than I, sorry, J less than K less than I, then you can have K less than I less than J, then you can have K less than J less than I, correct? So you can see that 6 such conditions are possible and this is just one of those 6 conditions. So this is one of the 6 conditions that you will get where your i, j and k are not equal. So 6 inequalities can exist. So saying this is as good as 6 times the required answer. That means my required answer will be 1 6th of 720. That is going to be 120. So option number A is the right option over here, absolutely Sondarya. So Vaishnavi is it clear now? Good. So guys this is actually an advanced level I would say this is not a main level question. It requires a deep understanding of sets and the loops within the loops. Anyways so we will now move on to the next question that is the 15th question now. The sequence x1, x2 till x50 yeah because they are symmetrical i, j, k they are all same series going from 1 to 10, 1 to 10, 1 to 10. Didn't you see that if you replace i, j and k with each other then equality still holds to be true each one of them is 120, 120, 120, 120, 120. So let's move on to the 15th question for the day. So the question says sequences x1, x2 till x50 has a property that for each k, xk is k less than the sum of the other 49 terms then the value of 96 x20. There should not be a comma over here, it should not be there. Okay, Mishis has given the first response. Yes, anyone else? Alright, so Atmish also thinks the same, somebody has come in the answer C okay. Alright, so mostly Ajanta is saying 15B okay, let's check whether B is the correct response or not. Okay, let us see what is the question saying. The question is saying that xk is k less than the sum of the other 49 terms. So can I say if I add k to this it will be the total sum minus xk right. So this is basically the sum of the other 49 terms okay, that is the sum of 49 other terms which means 2xk plus k is equal to s. Let us sum this expression from 1 to 50, so let's sum this expression both sides from k equal to 1 till 50 okay. So this will become 2 times summation of xk from 1 to 50 plus summation of k from k equal to 1 to 50 and this is just s 50 number of times, so it is 50s right. Now this is nothing but s because sum of xk from 1 to 50 is nothing but your s isn't it. So 2s plus summation of k from 1 to 50 we all know is n into n plus 1 by 2 and this is going to be 50s. So let us simplify this. So I can say 48s is equal to 25 times 51 okay and what do we need, we need her x20, we need 96 times x20 right. So how will I get that, I can get there that information from this expression once again. So I can say 2 times x20 is equal to s minus 20 that means my x20 could be written as half times s minus 20. So s is 25 into 51 by 48 so my x20 is going to be half times 25 into 51 by 48 minus 20 okay. Let us try to evaluate this but let us multiply with 96 also because it will make our life easy. So 96 x20 I am sure this will be 25 into 51 minus 20 into 48 yes. So this is going to give me 51 into 25 is going to give me 5 to 55 and this is going to be 102, 102 is 1275 minus 960 okay. This if I am not wrong is 315 so yes option number B is going to be the right option over here, option number B is going to be the right option over here. So the person who has given the right answer first is again wishes, by the way wishes do you know the source from where I have given these questions because you are getting almost every question correct. Now it has brought some doubt in my mind but really appreciate if you are getting it very good. Well done. I just kidding don't worry alright so we will move on to the next question. Let a0 be 0 and a n be 3 a n minus 1 plus 1 for n greater than equal to 1 find the remainder obtained when 820 is divided by 11 I think this is your question number 16 yes anyone sure sure guys again I am repeating please focus on chemistry most of you could have easily pulled out 99.5 percentile hydrochemistry been good and again repeating next 3 months chemistry is going to decide see maths and physics you more or less is all about your problem solving skill and you cannot make a colossal change or a very stark difference in the performance in math and physics but yes you can pull up chemistry within no time I have seen people you know pulling out pulling off a very very good rank just by focusing on chemistry for one or two months just keep solving questions there are very good books available which I have already told you the names NCRT is the first thing that you should know in and out you can't escape NCRT so please NCRT do it thoroughly do problems from books like Himanshu Pandey for organic Himanshu sir has given very good problem solving book on organic chemistry so please work out that book yes anyone alright let us start with a 2010 a 2010 we can say that it is 3 a 2009 plus one right similarly I can say a 2009 is three times a 2008 plus one plus one correct so if I'm not wrong it will give you three square a 2008 plus three plus one okay I've got a response from Sai so far okay let me stop here this is like a hint take this as a hint then some of you are just midway between problem solving I'm just giving you one more minute to see this okay so let's let's continue is it's not that difficult again we can say we can write this as 2003 into a 2007 plus one okay plus three plus one which is nothing but three cube a 2007 plus I think it'll give you three square right now can you see can you see the pattern that it's trying to show you if you do even further it'll give you three cube three a 2006 plus one plus three square three plus one that's going to be three to the power four a 2006 three to the power three three to the power two three plus one okay so if I go all the way till let's say three to the power 2010 a not so here what will I have here I will have 2009 2008 all the way till I reach three plus one is that correct now this term here would be zero because we know a not is zero so a not is zero so this term would be zero so it is as good as saying what will be the remainder when this is divided by 11 am I correct now this is a GP okay and the sum of this GP will be there are 2010 terms so three to the power 2010 minus one by two okay so how do you solve this thing how do you find out the remainder when three to the power 2010 minus one by two is divided by 11 any idea how to do this we can use the binomial okay so three to the power 2010 can we write it as a closest power plus minus one a power of 11 can we do that did you realize any power of n is it some plus minus one to a factor of 11 are you aware of any such power definitely two is not there three is also not there four will give you 81 that is also not there five okay if I split this as two plus one then okay we can get rid of that one anyways okay so okay let's let's follow the if you split this as two plus one then what happens let me just clear off this because you have already made use of this five is possible three to the power five is how much three to the power five is 243 right so 242 242 is going to be yes 242 is 0 by 11 right so what I'm going to do is I'm going to write it as 242 plus one and if I is it if I divide 2010 by five I'm going to get 402 okay minus one by two correct so that's going to give me 2000 242 to the power of 402 plus 402 C1 242 to the power of 401 and it continues all the way till I reach 402 C402 okay and that goes off with one right right so the next term left which is 402 C401 into 242 okay in fact when you divide by two when you divide by two you would still realize that each one of these terms can still take a factor of 11 okay so the answer is this will be completely a factor of 11 that means the remainder here was going to be 0 okay so option number a is going to be the right option over here let me not miss out a half over here right so this is the trick that you should always follow if you want to use your binomial theorem to find out there so try to pull out such a power of you know this term which is one plus minus of a factor of a multiple of 11 so in this case my five helped me out so this was my n actually so I wrote this as this and after that I use the binomial expansion and things after that are pretty clear great so we'll now move on to the next problem which is problem number 17 for the day so I think Saim here gave the right answer for the previous one well and say good next is we have a question suppose a1 a2 a3 till a 2012 our integers arranged on a circle right and each number is equal to the average of the two adjacent numbers if the sum of all even indexed numbers is 3003,018 okay then what is the sum of all the numbers then what is the sum of all the numbers I think super simple problem okay so all you started getting the answer d okay Gaurav says b let's check so we have numbers arranged in a circle so we have a1 a2 a3 a4 all the way till let's say 2010 a 2011 a 2012 okay now we have been given the expression that a2 is going to be a1 plus a3 by 2 right a3 is going to be a2 plus a4 by 2 if I continue in a similar way a 2012 is going to be a1 plus a 2011 by 2 and what else is given to me it's given to me the sum of all even indexed numbers that is a2 plus a4 all the way till a212 a2012 is 3018 now let's do one thing let's add let's write over here as a2 to a3 to a4 and so on till to a 2012 and here it will be I think every number let's say the a fourth term will be a3 plus a5 by 2 right so I think every term will appear twice over here except for a1 and a2 and the last two terms correct me if I'm wrong so a1 itself is a2 plus a 2012 by 2 right okay so let me write it over here a2 plus a 2012 by 2 so when you add all the terms on this side you are going to get two times guys you have to be very careful because some terms may be repeated some terms will not be repeated so when you're adding it you will be getting for example if you twice this up a2 to a4 etc you will be getting 3036 this will be 6036 okay and on this side you are going to get let me make some space over here so a1 plus a3 and you'll get everything twice one mistake one mistake just one second yeah so a1 plus a2 plus a3 and we'll go on all the way till a2011 again a1 and that will be this okay so what I did is I doubled up this and I started replacing 2 a2 with a1 plus a3 and 2 a3 with a2 plus a4 correct so if I keep doing like this you would realize that you will get twice of all these terms no you'll get all these terms over here I think I've missed out writing few of them so let me write a1 plus a3 this will be sorry 2 a4 2 a4 will be a3 plus a5 and a6 will be a5 plus a7 and so on and 2 a12 will be a2011 plus a21 and that is going to give me this so if I'm not mistaken I will be writing 2 times a1 plus a3 etc till a2011 as 6036 so cancel this so it will become 3018 so now a1 plus a3 plus a5 till a2011 was 3018 so when you add both of them you get a1 plus a2 all the way till a2012 as 6036 okay this was a simple question again I think the circular series was a little bit of confusing but never mind we could do it moving on to the next question that's the last question for the day find the sum of the series 9 by 5 square 3 1 sorry 2 1 13 by 5 cube 3 into 2 all the way till infinity again in the interest of time I'll be helping you with this so can I say I can write the terms in the numerator as 4r plus 1 of course r should be greater than equal to 2 here divided by 5 to the power r into rr minus 1 right this can be written as my rth term okay now what I will do is I will try to see whether I can apply my vn method so I can write 5 4r plus 1 as 5r minus I can see 5 over here so 5r and this r minus 1 okay if you split this up it becomes 5r by 5 to the power r into rr minus 1 minus r minus 1 by 5 to the power r rr minus 1 so this will get cancelled one of the factors of 5 will get cancelled rr will get cancelled and perfect I get this as a difference of two such expressions which are consecutive in r right now let's start putting the value of r as 2 3 and so on so when you start with 2 you get 1 by 5 okay in fact I can just write it on top over here so r is equal to 2 you'll get t2 the first term as in fact starts with the second term second term is going to be 1 by 5 minus 1 by 5 square into 2 then you'll have 1 by 5 square into 2 minus 1 by 5 cube into 3 and if you keep on going till infinity you'll realize all terms except the first term will remain first term will get cancelled so only the first term will remain which is option number c which is correct so the sum of this two infinite terms is going to be 1 by 5 okay so I think I had done this problem with you before also I think while we are solving for the first day main we did this problem anyways guys so I'll call it a day now thank you so much for coming online for this youtube live session and all the best for your boat practicals coming up thank you have a good night bye bye take care over and out from Centrum Academy