 Hi, and welcome to the session. I am Deepika here. Let's discuss a question. The sum of the third and seventh terms of an AP is 6 and their product is A. Find the sum of first 16 terms of the AP. We know that the n-th term of an AP with first term A and the common difference D is given by An is equal to A plus n minus 1 D and sum of first n terms of an AP is given by Sn is equal to n by 2 into 2A plus n minus 1 into D. So, let us use this key idea to solve up a question. So, let's start the solution. We know that n-th term of an AP An is equal to A plus n minus 1 D where A is the first term and D is the common difference. Therefore, A3 is equal to A plus 3 minus 1 D that is 2D and A7 is equal to A plus 60. Now, according to the question A7 is equal to 6 that is A plus 2D plus A plus 60 is equal to 6. This implies A plus 2A plus 8D is equal to 6. Divide this equation by 2 we get. A plus 4D is equal to 3. So, let us give this as equation number 1. Now, again according to the question and 7 term is 8 that is A3 into A7 is equal to 8 or we can say that A plus 2D into A plus 60 is equal to 8. This implies A square plus 6AD plus 2AD plus 12D square is equal to 8. Again, this implies A square plus 8AD plus 12D square is equal to 8. Now, let us give this equation as 2. Now, from equation 1 we have A plus 4D is equal to 3. This implies A is equal to 3 minus 4D. So, substitute this value of A in equation number 2. Substitute the above value of A is equal to 3 minus 4D in equation 2. We get this is equal to 3 minus 4D whole square plus 8 into 3 minus 4D into D plus 12D square is equal to 8. This implies 9 minus 24D plus 16D square plus 24 minus 32D, 24D minus 32D square plus 12D square minus 8 is equal to 0. So, on cancellation we get 1 minus 4D square is equal to 0. This implies 4D square is equal to 1 and this implies D square is equal to 1 by 4 is equal to plus minus 1 by 2. So, we have 2 values of D, D is equal to plus 1 by 2 and D is equal to minus 1 by 2. So, let us take 1 by 1 case 1, 1D is equal to 1 by 2. So, in this case A is equal to 3 minus 4D from 1. Therefore, this implies A is equal to 3 minus 4 into 1 by 2 which is equal to 1. Therefore, A is equal to 1. Now, A is equal to 1 and common difference D is equal to 1 by 2. So, our sum is sum of first n terms is given by S n is equal to n by 2 into 2A plus n minus 1D. So, at 16 we have to find the question. So, at 16 is equal to n is 16 here 16 by 2 into 2A 2 into 1 plus n minus 1 and these are 16, 16 minus 1 into 1 by 2. So, let us solve it. We get S 16 is equal to A 15 by 2. This implies S 16 is equal to 8 into 19 by 2. So, on cancellation we get S 16 that is sum of the first 16 terms is 76. So, this is a case when D is equal to 1 by 2. Let us consider the case when D is equal to minus 1 by 2 when D is equal to minus 1 by 2. We have A is equal to 3 minus 4D. This implies A is equal to 3 minus 4 into minus 1 by 2. So, on cancellation we get A is equal to 5 sum of first 16 terms that is S 16 is equal to n by 2 16 by 2 into 2A 2 into 5 plus n minus 1D that is 16 minus 1 into D in this case is minus 1 by 2. So, on solving we get this is equal to S 16 is equal to 8 10 minus 15 by 2. S 16 is equal to 8 into 20 minus 15 by 2. Therefore, S 16 is equal to 4 into 5 that is 20. So, we have two answers for the above question S 16 that is sum of first and terms is 20 when D is minus 1 by 2 and it is 76 when D is 1 by 2. So, this is our answer. I hope the question is clear to you. Bye and have a good day.