Let a be the first term and d be the common difference of the given A.P.,we have a = − d = − $\frac{11}{2}$ + 6 = $\frac{-11+12}{2}$ = $\frac{1}{2}$ Let n be the number of terms whose sum is – 25. ∴ $S_n$ = – 25 ⇒ $S_n$ = $\frac{n}{2}$ [2a + (n - 1) d] ⇒ - 25 = $\frac{n}{2}\left[-12+(n-1)\frac{1}{2}\right]$ ⇒ - 50 = - 12n + $\frac{n^2-n}{2}$ ⇒ - 100 = - 24n + $n^2$ - n ⇒ $n^2$ - 25n + 100 = 0 ⇒ $n^2$ - 20n - 5n + 100 = 0 ⇒ (n - 5) (n - 20) = 0 ⇒ n = 5 , 20 Hence either 5 terms or 20 terms are needed to give the sum – 25.