Let a be the first term and d be the common difference. We have a = 25, d = 22 – 25 = –3, $S_n$ = 116 Since $S_n$ = $\frac{n}{2}$ [2a + (n - 1) d] ∴ 116 = $\frac{n}{2}$ [50 + (n - 1) (- 3)] ⇒ 232 = 50n – $3n^2$ + 3n ⇒ $3n^2$ – 53n + 232 = 0 ⇒ $3n^2$ – 24n – 29n + 232 = 0 ⇒ (3n – 29) (n – 8) = 0 ⇒ n = $\frac{29}{3}$ , 8 Since n ≠ $\frac{29}{3}$ as n can not be in the fraction form ∴ n = 8 Hence, the last term $a_8$ = a + 7d = 25 + 7 (– 3) = 25 – 21 = 4.