 Hi and welcome to the session. I am Shashi and I am going to help you to solve the following question. Question is the 17th term of an AP exceeds its 10th term by 7. Find the common difference. Let us now start with the solution. We know nth term of an AP represented by a n is equal to a plus n minus 1 multiplied by d where a is the first term of the AP and d is the common difference between the two consecutive terms. Now we can write 17th term is equal to a plus 17 minus 1 multiplied by d or we can write a 17 is equal to a plus 16 d. Now similarly we can find the 10th term. So we can write 10th term that is a 10 is equal to a plus 10 minus 1 multiplied by d. This implies 10th term of the AP is equal to a plus 19. Now according to the question 17th term minus 10th term is equal to 7. Now we know 17th term is equal to a plus 16 d. So we can write a plus 16 d minus 10th term is equal to a plus 19. So we can write a plus 19 here is equal to 7. Now this implies a plus 16 d minus a minus 19 is equal to 7. Now this further implies a and a will get cancelled. So we get 17 is equal to 7. Now this implies d is equal to 7 divided by 7. So we get d is equal to 1. This common difference d is 1. So our required answer is 1. This completes the session. Hope you understood the session. Take care and goodbye.