 Hi and welcome to the session, I am Shashin and I am going to help you to solve the following question. Question is which term of the AP, 3, 15, 27, 39 will be 132 more than its 54th term? Let us now start with the solution, we know nth term of the AP is given by an 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 constitutive terms of the AP. Now the AP given to us in the question is 3, 15, 27, 39, clearly first term of the AP is equal to 3 and difference, common difference is equal to 15 minus 3 equal to 12, so we can d is equal to 12 and a is equal to 3. Now let us find out the 54th term of the AP, now we know 54th term of the AP is given by a plus 54 minus 1 multiplied by d, now value of a is equal to 3, so we will substitute 3 here and value of d is equal to 12, so we will substitute 12 in place of d, now simplifying we get 639, so 54th term of AP is 639, now let us assume that nth term of AP is 132 more than the 54th term, so we can write let nth term of AP be 132 more than 54th term that is a n is equal to 639 plus 132 or we can say nth term is equal to 771, now we also know that nth term is equal to a plus n minus 1 multiplied by d, now nth term is equal to 771, so we will substitute its value here is equal to 3 as a is equal to 3 and d is equal to 12, so we will substitute their corresponding values and get this equation, now simplifying we get 771 minus 3 is equal to 12 multiplied by n minus 1 this implies 768 is equal to 12 multiplied by n minus 1 or we can write 12 multiplied by n minus 1 is equal to 768, now this further implies n minus 1 is equal to 768 upon 12 which is equal to 64, so we get n minus 1 is equal to 64 or n is equal to 64 plus 1 equal to 65, so n is equal to 65, so our required answer is 65th term, 65th term will be 132 more than the 54th term, this is the required answer, hope you understood the session, take care and goodbye.