How many terms of the A.P. 25, 22, 19, …. are needed to give the sum 116 ? Also find the last term.

Let the number of terms required be n.

For the given A.P.,
a = 25, d = 22 - 25 = -3, S_n = 116

By formula, S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

$\Rightarrow 116 = \dfrac{n}{2}[2 \times 25 + (n - 1)(-3)] \\[1em] 116 \times 2 = n[50 - 3n + 3] \\[1em] 232 = n[53 - 3n] \\[1em] 232 = 53n - 3n^2 \\[1em] 3n^2 - 53n + 232 = 0 \\[1em] 3n^2 - 24n - 29n + 232 = 0 \\[1em] 3n(n - 8) - 29(n - 8) = 0 \\[1em] (3n - 29)(n - 8) = 0 \\[1em] 3n - 29 = 0 \text{ or } n - 8 = 0 \\[1em] n = \dfrac{29}{3} \text{ or } n = 8.$

Since, number of terms will be a natural number so n ≠ $\dfrac{29}{3}$

∴ n = 8.

By formula S_n = $\dfrac{n}{2}[a + l]$

⇒ 116 = $\dfrac{8}{2}[25 + l]$
⇒ 116 = 4[25 + l]
⇒ 29 = 25 + l
⇒ 29 - 25 = l
⇒ l = 4.

Hence, number of terms = 8 and the last term = 4.