 Hello and welcome to the session. The given question says the 7th term of an AP is 32 and the 13th term is 62, find the AP. Let's begin with the solution. So here we are given that the 7th term which is denoted by A7 is 32 and the 13th term is equal to 62. Now An is equal to A plus N minus 1 into D where A is the first term and N is the nth term and D is the common difference. So let us assume that the first term of the AP be A and let the common difference of the AP be D. So a required AP will be of the form, the first term will be A, second will be A plus D, third will be A plus 2D, fourth will be A plus 3D and so on with this term be A1, second term be A2, third term be A3, fourth term be A4 and so on. Now we are given that the 7th term of AP is 32. So by using this formula we have A plus 6D is equal to 32 and the 13th term is 62. So this implies A plus 12D is equal to 62. Let this be equation number 1 and this be equation number 2. Now let us subtract the second equation from the first one to get the value of D. We get here minus 6D is equal to minus 30 or D is equal to 5. Now substituting the value of D which is equal to 5, an equation number 1 to get the value of A we have A plus 6 into 5 is equal to 32 or we have A is equal to 32 minus 30 which is equal to 2. Therefore A is equal to 2 and D is equal to 5. Therefore the required AP, first term is A that is 2, second term is A plus D so 2 plus 5 is 7, third term is A plus 2D so we have 2 plus 2 into 5 which gives 12 then we have A plus 3D which gives 17 and so on. So this is the answer to the question. This all worked. Hope you understood it well. Bye and take care.