 Hello and welcome to the session. I am Asha and I am going to help you with the following question that says if the sum of certain number of terms of the AP 25, 22, 19 is 116, find the last term. Let us now begin with the solution and the given AP is 25, 22, 19 and so on. Now from this AP its first term is 25 and the common difference equal to 22 minus 25 which is equal to minus 3 and let the last term be equal to an. So an will be equal to a plus n minus 1 into d and also we are given that sum of n terms is equal to 116. So sum of n terms is denoted by n upon 2 to a plus n minus 1 into d. So this is the formula of sum of n terms. So this is equal to 116 and now we are equal to n upon 2 into 2 times of a a is 25 plus n minus 1 and d is minus 3 is equal to 116 or we have n upon 250 minus 3n plus 3 is equal to 116 or we have n upon 2 into 53 minus 3n is equal to 116. So this implies that 53n minus 3n square is equal to 232 or we have 3n square minus 53n plus 232 is equal to 0. Now by splitting the middle term this can further be written as 3n square minus 24n minus 29n plus 232 is equal to 0. Now taking 3n common from the first two terms and minus 29 from the last two terms it can further be written as 3n into n minus 8 minus 29 times n minus 8 is equal to 0 which further implies n minus 8 into 3n minus 29 is equal to 0. Now as we know if the product of two numbers is equal to 0 then at least one of them is 0 so this implies either minus 8 is equal to 0 or 3n minus 29 is equal to 0. So this implies n is equal to 8 or 29 upon 3. Now n cannot be 29 upon 3 therefore n is equal to 29 upon 3 is projected and therefore n is equal to 8. Now we have to find the last term so last term will be and that is equal to 8 plus 8 minus 1 into D 25 plus 8 minus 1 is 7 and D is minus 3 so this is equal to 25 plus 7 into minus 3 is minus 21 so this is equal to 25 minus 21 which is equal to 4. The last term which we have to find out is equal to 4. So this completes the session take care and have a good day.