 Hello and welcome to the session that has discussed the following question. It says the first term of an arithmetic progression AP is 26 and the nth term is 51 and the term in the 15th term of an AP. So let us now move on to the solution. We know that the nth term of an arithmetic progression that is AP is denoted by an and it is given by a plus n minus 1 into b where a is the first term of the AP and b is the common difference. Now we are given that the 5th term is 26 and 10th term is 51. So a5 is given by a plus n minus 1 where n is 5 so it is 5 minus 1 into b and we are given that a5 is 26. So we have a plus 4d is equal to 26 let us name this as 1 and we are given that the 10th term is 51. So 51 is a10 and it is given by a plus 10 minus 1 into b. So we have a10 is equal to 51 which is given by a plus 9d and this is equal to 51 let us name this as 2. Now we will solve these two linear equations for a and d. So subtract equation 2 from 1 so we have equation 1 is a plus 4d is equal to 26 and equation 2 is a plus 9d is equal to 51. Now we have to subtract so when we subtract sign will change plus a gets cancelled with minus a, 4d minus 9d is minus 5d and 56, 26 minus 51 is minus 25. So this implies d is equal to 25 by 5 minus sign gets cancelled both sides. So we have d is equal to 5. Now we will substitute the value of d in equation 1 to get the value of a so put is equal to 5 in 1. So we have a plus 4 into 5 is equal to 26. So this is equal to a plus 20 is equal to 26. So this implies a is equal to 26 minus 20 that implies a is equal to 6. So first term is 6 and the common difference is 5. Now we can easily find the 15th term of the arithmetic progression using this formula. So a15 is given by a plus 15 minus 1 into d now a is 6, 14 into d, e is 5 this is 6 plus 70 and it is equal to 76. Hence the 15th term of the arithmetic progression is 76. So this completes the question and the session by for now take care, have a good day.