 Hello and welcome to the session. Let us understand the following question which says, if the sum of the first n terms of an AP is 4n minus n squared, what is the first term that is s1? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the nth terms. Now let's proceed on to the solution. Sum of n terms of an AP is given to us as minus n squared. Therefore s1 is equal to, substituting n is equal to 1, in the given sum we get 4 multiplied by 1 minus 1 squared which is equal to 4 minus 1 which is equal to 3. Therefore first term that is s1 is equal to 3. Now sum of first two terms that is s2 is equal to, substituting n is equal to 2, in the given sum we get 4 multiplied by 2 minus 2 squared which is equal to 8 minus 4 which is equal to 4. Therefore s2 is equal to 4. Now let us find the second term. 2 is equal to n by 2 multiplied by 2a plus n minus 1d where n is equal to 2 and s2 is equal to 4 which is equal to 2 by 2 multiplied by 2 multiplied by a. That is our first term which is equal to 3 plus 2 minus 1 multiplied by d. Here we can see that we have only one unknown that is d. So 2 and 2 gets cancelled. 4 is equal to 6 plus d. This implies d is equal to 4 minus 6. This implies d is equal to minus 2. So we have our common difference is equal to minus 2. Therefore we have now first term is equal to 3. Sum of first two terms is equal to 4. Second term a2 is equal to minus s1 which is equal to 4 minus 3 which is equal to 1. Therefore second term a2 is equal to 1. Now let us find s3. s3 is equal to substituting n is equal to 3 in the given sum. We get 4 into 3 minus 3 square which is equal to 12 minus 9 which is equal to 3. Therefore s3 is equal to 3. Now we have to find the third term that is a3 which is equal to s3 minus s2 which is equal to 3 minus 4 equal to minus 1. Therefore third term a3 is equal to minus 1. Now we have to find the tenth term. So let us first find s9 and s8. s9 is equal to substituting n is equal to 9 in the given sum we get 4 into 9 minus 9 square which is equal to 36 minus 81 which is equal to minus 45. Therefore s9 is equal to minus 45. Now let us find s10. s10 is equal to again substituting n is equal to 10 in the given sum we get 4 into 10 minus 10 square which is equal to 40 minus 100 which is equal to minus 60. Therefore s10 is equal to minus 60 and hence the tenth term a10 is equal to s10 minus s9 which is equal to minus 60 minus minus 45 which is equal to minus 60 plus 45 which is equal to minus 15. Therefore a10 is equal to minus 15. Now we have to find the nth term. So let us find sn and sn minus 1. sn is given to us as 4n minus n square and sn minus 1 is equal to substituting n as n minus 1 we get 4 into n minus 1 minus n minus 1 the whole square which is equal to 4n minus 4 minus n square plus 1 minus 2n which is equal to 4n minus 4 minus n square minus 1 plus 2n which is equal to 6n minus 5 minus n square. Therefore an the nth term is equal to sn minus sn minus 1 which is equal to 4n minus n square minus 6n minus 5 minus n square which is equal to 4n minus n square. Now opening the brackets minus 6n plus 5 plus n square minus n square and plus n square gets cancelled. So we get 4n minus 6n is equal to minus 2n plus 5 which is equal to 5 minus 2n and hence the nth term an is equal to 5 minus 2n. Therefore the required answer is n minus s1 is equal to 3, s2 is equal to 4, a2 is equal to s2 minus s1 which is equal to 1, s3 is equal to 3, a3 is equal to s3 minus s2 which is equal to minus 1, a10 is equal to s10 minus s9 which is equal to minus 15 and an is equal to sn minus sn minus 1 which is equal to 5 minus 2n and this is our required answer. I hope you understood this question. That is all for this session. Bye and have a nice day.