 So hello everyone. In the last video, we have covered all the questions of exercise one of sequence and series chapter. So in this video, we will be covering the exercise two, we will discuss all the questions of this exercise. So let's begin with question number one. It is saying that the nth term of the series 25 plus 29 plus 33 plus dot dot dot and 3 plus 4 plus 6 plus 9 plus 13 plus dot dot dot are equal. Like the nth term of both the series are equal. Then we have to find the value of n. So one series is given here, this s1. Suppose we take first series as s1. This was 25 plus 29 plus 33 plus 37 plus dot dot dot. And one more series is given s2 as 3 plus 4 plus 6 plus 9 plus 13 plus dot dot dot. So clearly s1, series s1 is in AP with the common difference of 4. So we can write the nth term of this series as a plus n minus 1 into b. Where a is first term is 25 and the common difference is 4. So our nth term for this series will be 4n plus 21. And for this second series, if we see the difference of the terms like 4 minus 3, 1, 6 minus 4, 2, 9 minus 6, 3, 13 minus 9, 4, the difference of the terms are in AP. So for such type of series, we can find the nth term as a n square plus bn plus c. So this will be the nth term, nth term of the given series s2. Where a, b, c are some constants, we can find the value of a, b, c by putting the different values of n. So suppose we take n equal to 1, we will get t1 as a plus b plus c, right? And what is t1 in the series? In the series s2, it is equal to 3. Similarly, putting n equal to 2, we will get the second term which will be 4a plus 2b plus c is equal to 4. Because the second term in this series is 4. And putting n equal to 3, we will get 9a plus 3b plus c is equal to 6. So clearly, we are having three variables a, b and c. And we are having three equations, so we can find the value of a, b and c. So subtracting equation 1, suppose I am taking this as equation 2, this as 2 and this equation as 3. So subtracting equation 1 from equation 2, we will get 3a plus b is equal to 1. And subtracting 2 from 3, we will get 5a plus b is equal to 2. So again, subtracting, we will get 2a is equal to 1 or a is equal to 1 by 2. And putting the value of a in this equation, we will get 3 by 2 plus b is equal to 1. This will give b as minus 1 by 2. And putting the value of a and b in the first equation, we will get c is equal to 3. So nth term for this series, I am denoting it as Tn. So let me denote the first series nth term as Tn dash. So Tn dash was 4n plus 21. And Tn, nth term for the second series will be half n square minus half n plus 3. So as per question, both the nth terms are equal. So the Tn, this nth term is equal to nth term of the first series that is 4n plus 21. So solving this, we will get half n square minus 1 by 2n plus 3 is equal to 4n plus 21. So this is a quadratic in n, n square minus n plus 6 is equal to 4n plus 21. Or n square, sorry, this will be multiplied by 2. So this will be 8n and this will be, let me write once again, this 4n plus 21. So it will be multiplied by this 2. So this will be 8n plus 42. So n square minus 9n minus 36 is equal to 0. So further we can characterize this as n square minus 12n plus 3n minus 36 is equal to 0. Taking n common, it will be n minus 12 plus 3n minus 12 is equal to 0. From here it will be n plus 3 into n minus 12 equal to 0. From here we will get the two values of n, n is equal to minus 3 or n is equal to 12. Since n is the total number of terms, it cannot be negative and this value is rejected and this value will be accepted. So this will be our answer to this question. So let's see. So n equal to 12. So option b is correct for this question. Right? So let's move to the next question. So what is given? One series is given in the question and it is asking to find the arith term of the series. So one very like it seems to be very difficult combination of terms. So from exam point of view, we can cross shake with the options. We can cross shake with the options like four options are given and we have to find the arith term. So suppose I am taking option a. In option a, the arith term is given as 20 upon 5r plus 3. So let's put r equal to 3. If we put r equal to 3, we will have the third term as 20 upon 5 into 3 that will be 15 plus 3. That will be 20 upon 18 or 10 upon 9. So what is third term in our series? If we see the third term here, third term is 1 over 9. This is nothing but 10 upon 9. So we are finding that both the third terms like this third term, what we get from this arith term option a and the third term in the series both are equal. Hence option a is correct. So this approach we can like apply from exam point of view so that we can get the answers quickly. But suppose we go through our genuine method. So let's say the second method. The series is given as let me simplify and write this series as 4 5 upon 2 plus 20 upon 13 plus 10 upon 9 plus 20 upon 23 plus dot dot dot. So what I will do, I will try to make it simple like it is looking in a vague form. So I am writing one separate series taking the reciprocals of the terms. Like I will reciprocate the terms and I am writing the another series as 2 upon 5 plus 13 upon 20 plus 9 upon 10 plus 23 upon 20 plus dot dot dot. And now I will try to make the common denominator in all the terms. So I will multiply here by 4 numerator and denominator by 4. So it will become 8 upon 20. Since denominator is 20 here, let it be as it is 13 upon 20. Then I will multiply by 2 here 18 by 20 and 23 upon 20. Since 20 is already there, let it be as it is. So now we can see the numbers in the numerator are in AP. 8, 13, 18, 23 with the common difference of 5. So we can write the rth term for this series as 8 plus r minus 1 into 5 upon 20. So this will become 5r 8 minus 5. This will be plus 3. 5r plus 3 upon 20. But let me remind you this rth term is for h dash. This rth term is for h dash series. But in question, our original question was the s series. So rth term for s series will be equal to, we have to take reciprocal of this. So 20 upon 5r plus 3. 20 upon 5r plus 3, which we got in the first method itself, this tr. So 20 upon 5r plus 3. So this will be our answer. The rth term of given series will be option A. Hence option A is correct. Okay. So let's take the next question. Question number 3. What it is saying? In a certain AP, 5 times the 5th term is equal to 8 times the 8th term. Then its 13th term is. Okay. So it is given that 5 times 5th term of any AP is equal to 8 times the 8th term. And the question is asking to find the 13th term. What will be the 13th term of this AP? Okay. So 5 times 5th term. 5th term we can write as A plus 4d. Where A is the first term of the AP and D is the common difference. And this is equal to 8 times A plus 7d. We can write T 8, means 8th term as A plus 7d. From here we get 5A plus 20d is equal to 8A plus 56d. So taking this 5A to this side, it will be 3A plus 3060 is equal to 0. Taking 3 common, we will get A plus 12d is equal to 0 or A plus 12d is equal to 0. So this we get from the given condition. Now what is T 13th? Like what is 13th term? 13th term can be represented as A plus 12d. And we got the value of A plus 12d as 0. So 13th term of this series will be equal to 0. So option number A is correct for this question. Hope it is clear to all. So let's take the next question. Question number 4. If the 9th term of an AP is 0, the ratio of its 29th and 19th term is. So 9th term of an AP is 0. So this 9th term is 0. And we have to find the ratio of 29th term to 19th term. This we have to calculate. So 9th term is equal to 0. It means A plus 8d is equal to 0. 9th term we can write as A plus 8d where A is the first term and D is the common difference. So this condition is given. And now we have to find the ratio of 29th term at 19th term. So 29th term can be written as A plus 28d. And 19th term we can write as A plus 18d. And from here we can write A as minus 8d. So substituting the value of A in this, we can write it as A is equal to minus 8d plus 28d upon minus 8d. This is, this will become 20d upon 10d. So D will be cancelled out and this 20 will be divisible by 10 by 2. So the ratio of 29th term to 19th term will be 2 is to 1. So answer is option number B, right? So we will move to the next question. So moving to the next question, question number 5. It is saying that the, if the Pth, Qth and Rth term of an APR, AB and C respectively. So the value of this expression, one expression is given in the question and the, it is demanding the value of what will be the value of this expression. So let me write the given information. The Pth term, the Pth term is A, Qth term is B and the Rth term and the Rth term is C. Then we have to find the value of this expression A into Q minus R plus B into R minus P plus C into P minus Q. So let me simplify this expression. So this will be AQ minus AR plus BR minus BP plus CP minus CQ. So taking the Q terms, we can take common from, Q common from these two terms, it will be Q into A minus C. Similarly taking R common from these two terms, we will get R into B minus A. And similarly taking P common from here, we will get C minus B. So we have to simplify further this. So suppose I am taking this expression, sorry this term. So A minus C, A minus C, I am calculating only this expression. So what will be A? A will be the, A is the Pth term. So it will be A plus P minus 1 into D, where A is the first term and D is the common difference. So it will be A plus PB minus D. So okay, this minus, this was A and I forget to write this C. So C we can write as A plus R minus 1 into D, okay. So minus A minus R D plus D. So this D, this minus D will be cancelled, this minus A plus A will be cancelled. So this will be D into P minus R. So this term will become Q into A minus C what we got, P minus R into D. P minus R and let me write D here. So D Q into P minus R. Similarly what we get A minus C, we get A minus C as P minus R into D. Similarly B minus A will be, B minus A will be equal to D into B corresponds to the Qth term. So it will be Q minus and A corresponds to the Pth term. So it will be D into Q minus P and similarly C minus B will be, C corresponds to R and B means Q. So this will be D R into B minus A, B minus A is nothing but, okay D I have already mentioned. So Q minus P plus P and C minus B is D into R minus Q. So we will open the brackets now. So it will become D P Q minus D Q R plus D Q R minus D P R plus P D R minus P Q D. So this plus P Q D minus P Q D will be cancelled plus D Q R minus D Q R will be cancelled plus P D R minus P D R will be cancelled. So this will become 0. So finally this expression will turn out to be 0. So hence option C is correct for this question, option C. Okay. So this is done. Let's move to the next question. Question number 6. The sixth term of an AP is equal to 2. The value of common difference of AP which makes the product A1 into A4 into A5 list is given by. So A1 into A4 into A5 like first term, the product of first term, fourth term and fifth term. The question is asking that this product's value will be list then what will be the value of common difference. And the given information is that the sixth term of AP is equal to 2. So this sixth term of AP is 2. So sixth term we can write as A plus 5 D is equal to 2. And we have to find the value of common difference which makes this product A1 A4 A5 list. Okay. So this product what is given A1 into A4 into A5. Okay. A1 is the first term. So this sixth term we will write as A1 plus 5 D. So A1 into A4 A4 will be A1 plus 3 D. And this fifth term will be A1 plus 4 D. So from here we get A1 as 2 minus 5 D. So substituting the value of A1 2 minus 5 D 2 minus 5 D plus 3 D. So this will be 2 minus 2 D. And this will be 2 minus 5 D plus 4 D. Then this will be 2 minus D. So we can see this product is a function in D. This product is a function in D. And this will be actually a cubic function. This will be cubic function in D. So we can simplify it and we can write it in simplest form. Like 2 4 to the 4 minus minus 4 D and 10 D minus 14 D plus 10 D square. Again it will be multiplied by 2 minus D. So this will be 4 to 8. Then 4 minus 4 D minus 4 D and this minus 18 D. So let me write it differently. Minus 4 D minus 28 D plus 14 D square plus 20 D square and minus 10 D cube. So it will be minus 10 D cube plus 34 D square and then minus 32 D plus 8. So this is our product which we are representing as F of D. Since it is a cubic function in D and the leading factor here if we see, the leading factor of this cubic function is negative. So for any cubic if you see, if we plot the graph of any cubic function, Y is equal to F of D. If we plot the graph of any cubic function with a negative leading coefficient, so it will start from the like plus infinity. The graph will somewhat look like this. It will start from, it will go from minus infinity to plus infinity. But the question is saying that the product should be minimum. So actually it should not, the minimum value of this function will be actually minus infinity. So there should be some correction in this question. Actually the question is trying to say that the product A1, A4, A5, where it is taking the minimum local minima. So it is trying to find the value of that D from at which the function is taking the local minimum value. So the question is trying to find the value of this D. Since if you see the least value of this function it will be minus infinity. So similarly we can, we have to find this value of D where the function is giving local minimum value. So how to find? This is our function in D. So we will differentiate this function. We will differentiate this function and we will make it as 0. The slope of tangent will be 0 at the point of local minimum or maximum. So I am differentiating this function with respect to D. It will be minus 30 D square plus 68 D minus 32 and we will put this as 0. So it will become taking 2 common minus 15 D square plus 34 D minus 16 is equal to 0. Taking minus common it will be 15 D square minus 34 D plus 16 is equal to 0. From here we will get 2 values of D since it is quadratic in D. So 15 into 16 will be 150 plus 19, 240. So we can break it as 24 and 10. So 15 D square minus 24 D minus 10 D plus 16 is equal to 0. Taking 3 D common we will get 5 D minus 8 and minus 2 if we take 2 common it will be 5 D minus 8 is equal to 0. So 5 D minus 8 into 3 D minus 2 is equal to 0. From here we get D is equal to 2 by 3 or D is equal to 8 by 5. So if we observe this graph 2 points are coming. So one of the point will be point of local minimum and one of the point will be point of local maximum. So if we see if we further differentiate this function like we have to take this was our F dash D. So further differentiating this F double dash D we will get minus 60 D plus 68. And what will be the value of double derivative at D equal to 2 by 3? F double dash 2 by 3 will be minus 60 into 2 by 3 plus 68. This will be 320 minus 40 plus 68 that is plus 28. So this 2 by 3 will be the point of local minimum since F double dash at that point is coming out to be positive. Similarly if we see F double dash at 8 by 5 we will get minus 60 into 8 by 5 plus 68 which will be 12. 96 plus 68 that is minus 28. So this 8 by 5 will be the value of 8 by 5 at D equal to 8 by 5 the function is taking the local maximum value. So actually this is our 2 by 3 point this point is 2 by 3 comma 0. And this point where the function is taking local maximum value this is 8 by 5 comma 0 right. So our answer will be 2 by 3 but as informed you all like there is slight correction in this question. It will not be the least value this product is not least because the product is product is going to minus negative. So least value will be minus negative but the question is trying to find the value of D where the function is taking local minimum. So our answer will be 2 by 3 so option C is correct. Okay so moving ahead we will take the next question. What it is saying the sum of first two in terms of an A B is alpha and the sum of next in terms is beta then its common difference is okay. So suppose one series is given A1 plus A2 plus A3 up to two in terms so this plus two in terms. So sum of first two in terms of an A B is alpha and next in terms is beta okay. So let me extend this series as up to three in the three in terms. So next term will be A2n plus one then A2n plus two it is going up to A3n three in terms. So question is saying the sum of first two in terms sum of first two in terms is alpha and sum of next next in terms. So sum from 2n plus 1th term to 3nth term is beta. So what will be our sum to three in terms? What will be sum to our three in terms? It will be alpha plus beta right. So this will be our sum to three in terms. What it is asking it's asking to its common difference. Question is asking to find the value of D. Question is asking to find the value of D means common difference okay. So two informations are given sum of two in terms is alpha and sum of three in terms is alpha plus beta and we have to find the value of D. So sum of two in terms we can write as 2n by 2 and 2A plus 2n minus 1 into D. And we can write the sum of three in terms as 3n by 2 2A plus 3n minus 1 into D. And this sum of two in terms is given as alpha and sum of three in terms is given as alpha plus beta. So let us simplify this. So these two and these two will get cancelled out. It will become 2an plus 2n minus 1 Dn is equal to alpha. And from here we get this two will be cancelled out. So 3an plus 3n D by 2 into 3n minus 1 is equal to alpha plus beta. Now we have to find the value of D. We can do one thing we can multiply this equation by three and this equation by two. So that this term will become 6an and this will also become 6an and we can further subtract first equation from second and we will find the value of D. So let us move with the same approach. So multiplying by three this will give 6an plus 3n D 2n minus 1 is equal to 3alpha and this will become 6an plus 3n D into 3n minus 1 is equal to 2alpha plus 2 beta. Suppose this take this equation as one, this equation as two. So subtracting equation one from two, subtracting equation one from two we will get 6an plus 3n D 3n minus 1 minus 6an minus 3n D 2n minus 1 is equal to 2alpha plus 2 beta minus 3alpha. So this 6n minus 6n we will take 3n D common. This will become 3n minus 1 minus 2n plus 1 is equal to 2 beta minus alpha. So we have to find the value of D. So D will be ok and this plus 1 minus 1 will be cancelled out this will remain only n. So this 2 beta minus alpha upon 3n and this n 3n square. So this will be the value of common difference 2 beta minus alpha upon 3n square. So option B, option B is correct. So we are done with this question number 7. So let us go move ahead and see the question number 8. What it is saying? The sum of three terms in AP is minus 3 and their product is 8. Then the sum of squares of the numbers. Then the sum of the squares of the numbers ok. So whenever the question is saying to take three, sum of three numbers are given or so we normally assume the terms as A minus D, A and A plus D. Because while taking some this D plus D and minus D will get cancelled out and we can easily find the value of A. So sum of these three terms this A minus D plus A plus A plus D is equal to minus 3. So this plus D minus D will get cancelled out. This 3A is equal to minus 3 and from here we get A is equal to minus 1. We get A is equal to minus 1 and their product is also given. So what will be product A minus D into A into A plus D and product is given as 8. So put the value of minus 1. So this will be minus 1 minus D minus 1 minus 1 plus D is equal to 8. So taking minus common from this it will become D minus 1 sorry minus means plus and plus D plus 1. D plus 1 into D minus 1 is equal to 8. So this will become D square minus 1 is equal to 8. So D square is equal to 9. From here we get two values of D plus minus 3. We get two values of D. So our AP will be minus 1 when D equal to we take D equal to 3. Then our AP will become minus 1 minus 3 comma minus 1 then minus 1 plus 3 it will minus 4 minus 1 and 2. And when we take D equal to minus 3 same AP will come but in reverse order that will be 2 minus 1 and minus 4. So the question is asking to find the sum of square of the numbers. So we have to take the square of these three numbers. So minus 4 is square plus minus 1 is square plus 2 is square. So it will become 16 plus 1 plus 4. This will be 21. So option D is correct. Option D is correct for this question. So it's done. Question number 8 is completed. So let's go ahead with next question. Question number 8 sorry question number 9. What it is saying. Let S and denote the sum of in terms of an AP. If sum of 2 in terms is equal to 3 times sum of in terms then the ratio of S 3 and 2 SN is equal to. Okay so it is given that sum of 2 in terms. Sum of 2 in terms is equal to 3 times sum of n terms. And we have to find the ratio of sum of 3 in terms to sum of n terms. So we can write the sum of 2 in terms as 2n by 2. 2A plus 2n minus 1 into D is equal to 3 times sum of n terms. Sum of n terms will be n by 2 2A plus n minus 1 into D. So this n by 2 this n by 2 will be cancelled. This will become 4A 4A plus 2D into 2n minus 1 is equal to 6A 6A plus 3D into n minus 1. From here we get 2A 2A sorry let me write separately. So bringing 4A to this side it will become 2A is equal to and taking D terms to that side it will become 2D 2n minus 1 minus 3D n minus 1. So further we can simplify anything. This will be 4Dn minus 2D minus 3Dn plus 3D is equal to 2A. So this will become Dn minus D. 4Dn minus 3Dn it will be Dn plus 3D minus 2D it will be plus 2. So Dn plus D is equal to 2A or taking D as common D into n plus 1 is equal to 2A. Let's see whether we can use this relation here. So question is asking to find sum of 3 in terms to sum of n terms. So it will become sum of 3 in terms will be 3n by 2 2A plus 3n minus 1 into D upon what will be the sum of n terms? Sum of n terms will be n by 2 2A plus n minus 1 into D. So this n by 2 n by 2 will be cancelled out. We can write D into n plus 1 in place of 2A. So 3 into 2A we can write D into n plus 1 plus 3n minus 1 into D and this will be D into n plus 1 plus n minus 1 into D. So 3 into if we take D common here we will get n plus 1 plus 3n minus 1 upon again taking D common here it will be n plus 1 plus n minus 1. So plus 1 minus 1 will be cancelled out. This D and D will get cancelled. This plus 1 minus 1 will get cancelled. It will become 3 into 4n 3n plus n 4n and in denominator we are having 2n. So this will be cancelled 2 and this will become 6. 3 into 2 will be 6. So the ratio of sum of 3n terms to the sum of n terms will be equal to 6. So option B will be correct for this question. So 6 will be correct answer. So moving ahead let's take the next question, question number 10. The sum of products of 10 numbers plus minus 1 plus minus 2 plus minus 3 plus minus 4 plus minus 5 taking 2 at a time is. So it is asking sum of products of 10 numbers. Sum of product of 10 numbers. Numbers are given as plus minus 1 plus minus 2 plus minus 3 plus minus 4 and plus minus 5. So question is saying that plus minus 1 like plus 1 into minus 1 what it will be? It will be minus 1. Similarly plus 2 and minus 2 product of plus 2 and minus 2 will be minus 4 plus 3 into minus 3 will be minus 9 plus minus 16 plus minus 25. So this will be 1 plus 4 5 9 plus 5 16 16 plus 16 32 no 1 plus 4 5 9 plus 5 14 14 plus 16 30 30 plus 25 55 with the negative side. Okay we can also we could have also approached this question as like we can write this as minus 1 square plus 2 square. I have taken minus common from all. So it will become 1 square plus 2 square plus 3 square plus 4 square plus 5 square and anyhow we know the sum of n terms is squared like sum of r square from r going from 1 to n. So what we know the value of that will be n into n plus 1 into 2 n plus 1 upon 6. So what is n here? n is 5 here. So let's put 5 so 5 into 6 into 2 5 10 11 upon 6. So 6 6 will be cancelled it will be minus 55. Then we could have approached this question with this logic also and since this is a only 5 terms are there so we could do this manually also. But if the sums would have been like many terms would have been involved then this approach would be more beneficial to us. This is more logical to apply this formula of sum of sum of the squares of the first and natural numbers this formula which we have derived in our class also. So option number c is correct option c is correct for this question. So I don't know I think one more question is there in this exercise. Okay this is the last question I think this is question number 11 and this is the last question of this exercise. So what it is saying if a1 a2 a3 dot dot dot up to n are in AP where ai is greater than 0. So all the terms of this AP are positive. The value of this series question is asking to find the value of this series. Series is given as 1 upon root a1 plus root a2 plus 1 upon root a2 plus root a3 plus dot dot dot dot dot up to 1 upon root an minus 1 plus root of an. So we have to find the value of this series. So we can do one thing we can multiply and divide the term by the conjugate of the denominator like we can write this as root of a1 minus root of a2 upon root of a1 plus root of a2 into root of a1 minus root of a2. Similarly we have to do this with all the terms so it will become root a2 minus root a3 upon and this will become actually if you see this will be a plus b and a minus b. So it will be actually a squared minus b squared. So I am writing here only that this denominator will actually become a1 minus a2. I am writing here only a1 minus a2. So this will become a1 minus a2. Similarly this term will become the denominator of this term will become a2 minus a3 plus root a3 minus root a4 upon a3 minus a4 and it will continue up to this last term which will be nothing but root of root of an minus 1 minus of root an upon an minus 1 minus an. If we observe this term this thing this thing in denominator a1 minus a2 a2 minus a3 a3 minus a4 this is nothing but the difference of two consecutive terms. And in AP the difference of two consecutive terms is dA actually. The only difference is it will be minus d right since it is a1 minus a2. So if it would have been a2 minus a1 so that would have been like plus d but here it will become minus d if we take it as an increasing AP. So let me write it as a upon a upon minus d. So this will become root a1 minus root a2 plus root a2 minus root a3 plus root a3 minus root a4 dot dot dot dot. This will continue up to plus root of a minus 1 minus of root of a1. So if we see this a1 sorry minus a2 plus a2 minus root a3 plus root a3 minus root a4 plus root a4 it will be cancelled. And what we will be left out with we will be left out with root of a1 minus root of an upon d upon minus d. And what we can write this this will become root a1 root a1 minus root an. And we can see we can write this as this d as or we can do one thing let me do in this way. Let me multiply this numerator and denominator by root of a1 plus root an upon root a1 plus root an. So it will become a1 the numerator will become a1 minus an upon this d d we can write as no we are getting stuck somewhere this a1 minus a1 minus an upon d this root a1 plus root an. Okay this will be okay this a1 minus an we can write it as we can write it as minus and d right this may be written as n minus 1 d right n minus 1 into d. Because the gap is gap between a1 and a1 a1 and an is n minus 1. So this difference will be n minus 1 into d okay so we can write it as so this d and this d will be get cancelled out it will be n minus 1 upon root a1 plus root an. So this will be answer to this question because this difference a1 difference between a1 and an is like n minus 1 gaps are created n minus 1 gaps are created and it will be the difference between each gap is d. So total difference will be n minus 1 into d. So this d and this d will be cancelled out it will be n minus 1 upon root a1 plus root an. So hope this is clear to everyone so we are done with this exercise in the coming video we will be taking the next exercise that will be exercise number three. So up to then take care goodbye.