 Hello and welcome to the session. I am Aasha and I am going to help you with the following question which says, let the sum of N to N, 3 N terms of an AP, be S1, S2 and S3 respectively, show that S3 is equal to 3 times of S2 minus S1. Let us now begin with the solution and let the first term of an AP is equal to A difference is equal to D. We have to show that it is equal to 3 times of S2 minus S1. Now sum of N terms S1 is equal to N upon 2 minus 1 into D, 2 is equal to 2N upon 2 into 2A plus 2N minus 1 into D. Sum of 3 N terms is equal to 3N upon 2 into 2A minus 1 into D. Now first let us find S2 minus S1, S2 is N into 2A plus 2N minus 1 into D minus S1 is minus 1 into D. So this is further equal to N is ND minus, now solving this we have AN minus N is equal to 2 into D plus N upon 2 minus AN is AN. Now solving this 2 we have N upon 2 N square D, now solving this 2 we have minus 1 upon 2 ND. So this is further equal to N, keep in common. So here we have 2A into 2A plus minus 1 into D. Now multiplying the numerator and denominator by 3 we have 3N upon 6 into 2A AN minus 1 into D and now since S3 is 3N upon 2 into 2A plus 3N minus 1 into D this can be written as S1 into D S3. So we have 1 upon 3 S1 is equal to 1 upon 3 S3 we can say that S3 is equal to 3 times of S2 minus S1, 3 times of S2 minus S1 where S1, S2 and S3 are the sum of N to N and 3N terms respectively often AP. So this completes the session take care and have a good day.