 Hello and welcome to the session. Let us understand the following question today. In an AP given A is equal to 3, N is equal to 8, S is equal to 192, find D. Now let us hide the solution. We know S is equal to N by 2 multiplied by 2A plus N minus 1B. Now substituting the value of A, N and S we get, S is given to us as 192 which is equal to 8 by 2 2 multiplied by 3 plus 8 minus 1D. This implies this 2 goes here so we get 384 is equal to 8 multiplied by 6 plus 7D. This implies 384 is equal to 48 plus 56D. This implies 56D is equal to 384 minus 48. This implies 56D is equal to 336. This implies D is equal to 336 divided by 56. Now it gets cancelled by 2 so we get here 168 and here we get 28. Now again it gets cancelled by 2 so we get here 84 and similarly here we get 14. Again it gets cancelled by 2 so we get here 42 and we get here 7. Now 76A is 42 therefore which implies D is equal to 6. Hence D is equal to 6 is our answer. I hope you understood the question. Bye and have a nice day.