 Hi and welcome to the session. Let's work out the following question. The question says find the number of terms of AP, 54, 51, 48 and so on, so that their sum is 513. So let us start with the solution to this question. We have first term that is A is equal to 54. Common difference that the small d is equal to 51 minus 54. That is same as 48 minus 51 that is equal to minus 3 and it's given to us that sum of n terms is 513 but we also know that sum of n terms is equal to n by 2 into 2A plus n minus 1 into d. So now putting down the values we have 513 is equal to n by 2 into 2 into 54 that is the first term plus n minus 1 into d that is minus 3 or 513 into 2 is equal to n into 108 plus n minus 1 into minus 3 or 1026 is equal to n into 3 into 36 minus n minus 1 or 1026 divided by 3 is equal to n into 36 minus n minus 1 or 342 is equal to 36n minus n square plus n or 342 is equal to 36n or we can say 37n minus n square or n square minus 37n plus 342 is equal to 0. Now factorizing this we get n square minus 18n minus 19n plus 342 is equal to 0 or n into n minus 18 minus 19 into n minus 18 is equal to 0. This implies that n minus 18 into n minus 19 is equal to 0 so this implies n is equal to 18 or 19. Now here 19th term that is t19 is equal to A plus now n will be 19 so n minus 1 into d putting down the values we have 54 plus 19 minus 1 into minus 3 that is equal to 54 minus 54 that is equal to 0. Now we see here that 19th term is 0 so sum of either 18 terms or 19 terms is 513. So this is our solution to this question I hope that you understood the solution and enjoyed the session have a good day.