 Hello and welcome to the session. In this session we discussed the following question that says how many terms of the series 54, 51, 48 and so on we take in so that the sum is 513. So for this we must know what is the sum of n terms of an AP it is given by Sn and this is equal to n upon 2 this whole into 2A plus n minus 1 into d. When this A is the first term of the AP, d is the common difference of the AP. This is the key idea to be used in this question. Let's move on to the solution now. The given series is 54, 51, 48 and so on where the first term A is 54 and the common difference d is equal to 51 minus 54 which is equal to minus 3. It's given that the sum of n terms of the AP that is Sn equal to 513. Now we know that Sn is given by n upon 2 this whole into 2A plus n minus 1 into d. So Sn is equal to n upon 2 whole into 2A plus n minus 1 into d and this is equal to 513. Now we put the values for A and d that is A is 54 and d is minus 3. So we get n upon 2 whole into 2 into 54 plus n minus 1 into minus 3 is equal to 513. This further gives us n upon 2 whole into 108 minus 3 into n minus 1 is equal to 513. From here we get n upon 2 whole into 108 minus 3n plus 3 is equal to 513 which further gives us n upon 2 whole into 111 minus 3n is equal to 513. This gives us 111 minus 3n whole into n is equal to 1026. From here we get 3n square minus 111n plus 1026 is equal to 0. Now we take 3 common inside the bracket we have n square minus 37n plus 342 this is equal to 0. So we get n square minus 37n plus 342 is equal to 0. Now solving this quadratic equation by splitting the middle term we get n minus 18 this into n minus 19 is equal to 0. So this means n minus 18 equal to 0 or n minus 19 equal to 0 which gives us n equal to 18 or n equal to 19. So this means the given AP has either 18 terms or 19 terms. Now let's find out the 19th term of the AP which is given by a19 and this is equal to a plus 19 minus 1 into d. Putting the values of a and d we get 54 plus 18 into minus 3 that is equal to 54 minus 54 equal to 0. So we get a19 is equal to 0 that is the 19th term of the AP is 0. Thus we say the sum of 18 terms as well as 19 terms is 513. So with this we complete this session hope you have understood the solution of this question.