In an A.P. the first term is 25, n^th term is -17 and the sum of n terms is 132. Find n and the common difference.

Given,

a = 25, a_n = -17 and S = 132

⇒ a + (n - 1)d = -17

⇒ 25 + (n - 1)d = -17

⇒ (n - 1)d = -42 ……..(i)

We know that,

$\Rightarrow S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(2 \times 25 + (-42)) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(50 - 42) \\[1em] 132 = \dfrac{8n}{2} \\[1em] 4n = 132 \\[1em] n = 33.$

Substituting value of n in (i),

⇒ (33 - 1)d = -42

⇒ 32d = -42

⇒ d = -$\dfrac{42}{32} = -\dfrac{21}{16}$

Hence, n = 33 and d = $-\dfrac{21}{16}$.