 Hello and welcome to the session. In this session, we discussed the following question which says for a given AP, the sum of its N to N, 3 N terms are S1, S2, S3 respectively, prove that S3 is equal to 3 into S2 minus S1. So first let's see what is the sum of N terms of an AP. It is given by Sn and this is equal to N upon 2 into 2A plus N minus 1 into D where this A is the first term of the AP. D is the common difference. This is the key idea to be used in this question. Let's see the solution now. We are given that the sum of the N terms of an AP is equal to S1 then the sum of 2N terms of an AP is given as S2 then sum of 3N terms of an AP is given as S3. And we need to show that S3 is equal to 3 into S2 minus S1. Now first we assume that A be the first term of the given AP and let D be the common difference of the given AP. So now as S1 is the sum of N terms, so S1 is given by N upon 2 into 2A plus N minus 1 into D. We take this as equation 1 then S2 that is the sum of 2N terms would be given by replacing N by 2N in S1. So 2N upon 2 into 2A plus 2N minus 1 into D let this be equation 2 then we have S3 it is the sum of 3N terms of an AP. So this is equal to 3N upon 2 into 2A plus 3N minus 1 into D. That is we have got this by replacing N by 3N in equation 1. So let this be equation 3. We need to show that S3 is equal to 3 into S2 minus S1. So first let's find out S2 minus S1 this would be equal to 2N upon 2 into 2A plus 2N minus 1 into D minus N upon 2 into 2A plus N minus 1 into D. Now we take N upon 2 common. So inside we have 2 into 2A plus 2N minus 1 into D minus 2A plus N minus 1 into D. Thus we have S2 minus S1 is equal to N upon 2 into 4A plus 2D into 2N minus 1 minus 2A minus N minus 1 into D. So from here we have S2 minus S1 is equal to N upon 2 into 4A minus 2A is 2A plus 2D into 2N is 4ND 2D into minus 1 is minus 2D then minus N into D ND plus D. So from here we have S2 minus S1 is equal to N upon 2 into 2A plus 3ND minus D that is S2 minus S1 is equal to N upon 2 into 2A plus 3N minus 1 into D. Next let's find out 3 into S2 minus S1 so this would be equal to 3N upon 2 into 2A plus 3N minus 1 into D. Now as you can see that we got S3 equal to 3N upon 2 into 2A plus 3N minus 1 into D that is we have now got 3 into S2 minus S1 is equal to S3 that is using equation 3. So hence we have proved that S3 is equal to 3 into S2 minus S1 so this completes the session hope you have understood the solution for this question.