 Hello and welcome to the session. Let us understand the following problem today. In an AP given A12 is equal to 37, D is equal to 3, find A and S12. Now let us write the solution. We know that A n is equal to A plus n minus 1 D, so which is calculated as A12 is equal to A plus 12 minus 1 multiplied by 3. A12 is given to us 37, so 37 is equal to A plus 11 into 3. Now here there is one unknown that is A. So A which implies A is equal to 37 minus 33 which implies A is equal to 4. Now using this value of A let us find the sum that is S12 which is equal to n by 2 multiplied by 2A plus n minus 1 D which is equal to n by 2 multiplied by 4 plus 12 minus 1 D is 3 which is equal to this gets cancelled by 6, so 6 multiplied by 8 plus 11 into 3 which is equal to 6 multiplied by 8 plus 33 which is equal to 6 multiplied by 41 which is equal to 246 hence A is equal to 4 and S12 is equal to 246 our required answer. I hope you understood the question by and have a nice day.