 Hello and welcome to the session. In this session, we will discuss a question which says that if the sum of first p terms of an a p is a p square plus b p, find its common difference. And the options are option a, a, option b, 2 a plus 2 b over p minus 1, option c, 3 a and option d, a plus b. Now before starting the solution of this question, you should know a result. And that is the sum of n terms in n a p is given by s n is equal to n by 2 into 2 a plus n minus 1 the whole into d where the common difference. Now this result will work out as a key idea, this question. And now we will start with the solution. Now given the sum of first p terms in an a p is a p square plus b p. So given the sum of p terms of an a p a p square plus b p. Now using this formula which is given in the key idea of an a p is equal to p by 2 into 2 a plus p minus 1 the whole into d. Now this is also the sum of first p terms of an a p. So this implies a p square plus b p is equal to p by 2 into 2 a plus p minus 1 the whole into d. Now this implies taking p common within brackets it will be a p plus b is equal to p by 2 into 2 a plus p minus 1 the whole into d the whole. Now where p will be cancelled with p. So this implies on first multiplying 2 a p plus 2 b is equal to 2 a plus p minus 1 the whole into d. However this implies plus 2 b minus 2 a is equal to p minus 1 the whole into d. Which implies from these two terms it is common and written like this p minus 1 plus 2 b is equal to p minus 1 the whole into d. This implies on the right of each term by p minus 1 it will be 2 a plus 2 b over p minus 1 is equal to d. Therefore d is equal to 2 a plus 2 b over p minus 1. Therefore the answer of this question is option b a plus 2 b over p minus 1. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.