Question

if the sum of a certain number of terms starting from first , of an ap 25,22,19....,is 116. find the last term

Answer 1

Hope these helps..

n must always be a whole number, that's why n=29/3 is ignored

Answer 2

AnswEr:

Let the sum of n terms of A.P . 25, 22, 19 , ... be 116.

Then,

$\\ \qquad \tt116 = \frac{n}{2} \: [2 \times 25 + (n - 1) \times ( - 3)] \\ \\ \\ \implies \tt \: 232 = n \: ( - 3n + 53) \\ \\ \\ \implies \tt3 {n}^{2} - 53n + 232 = 0 \\ \\ \\ \implies \tt \: 3 {n}^{2} - 24n - 29n + 232 = 0 \\ \\ \\ \implies \tt \: (n - 8)(3n - 29) = 0 \\ \\ \\ \implies \tt \blue {n = 8} \\ \\ \\ \therefore \: \: \rm \: a_g = 25 + 7 \times ( - 3) = 4$