 Hello and welcome to the session. Let us understand the following question today. In an AP given A is equal to 7, A13 is equal to 35, find A and S13. Now, let us write the solution. Now, we know that AM is equal to A plus N minus 1D. Now, substitute all the values that are given to us. AM is A13 which is equal to A that is 7 plus N is equal to 13 here minus 1D we have to find out. And A13 is given to us 35, so 35 is equal to 7 plus 12D. Now, solving it further it implies 35 plus is equal to 7 plus 12D which implies 12D is equal to 35 minus 7 which implies 12D is equal to 28 which implies D is equal to 28 by 12 which is equal to 7 by 3. Therefore, D is equal to 7 by 3. Now, we will find S13. We know that SN is equal to N by 2 multiplied by A plus AM and for S13 we can write S13 is equal to N by 2 multiplied by A plus A13. So, substituting all the values we get S13 is equal to 13 by 2 multiplied by 7 plus 35. Now, solving this we get which is equal to 13 by 2 multiplied by 42. Now, here this gets cancelled with this by 21 so we get which is equal to 273. Therefore, D is equal to 7 by 3 and S13 is equal to 273 is our required answer. I hope you understood the question. Bye and have a nice day.