In an A.P., given a_n = 4, d = 2, S_n = -14, find n and a.

Given,

a_n = 4

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 4 = a + 2(n - 1)

⇒ 4 = a + 2n - 2

⇒ a + 2n = 4 + 2

⇒ a + 2n = 6

⇒ a = 6 - 2n ……..(1)

By formula,

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

Given,

S_n = -14

Substituting values we get :

$\Rightarrow \dfrac{n}{2}[2 \times (6 - 2n) + 2(n - 1)] = -14 \\[1em] \Rightarrow \dfrac{n}{2}[12 - 4n + 2n - 2] = -14 \\[1em] \Rightarrow \dfrac{n}{2}[10 - 2n] = -14 \\[1em] \Rightarrow 10n - 2n^2 = -28 \\[1em] \Rightarrow 2n^2 - 10n - 28 = 0 \\[1em] \Rightarrow 2(n^2 - 5n - 14) = 0 \\[1em] \Rightarrow n^2 - 5n - 14 = 0 \\[1em] \Rightarrow n^2 - 7n + 2n - 14 = 0 \\[1em] \Rightarrow n(n - 7) + 2(n - 7) = 0 \\[1em] \Rightarrow (n + 2)(n - 7) = 0 \\[1em] \Rightarrow n + 2 = 0 \text{ or } n - 7 = 0 \\[1em] \Rightarrow n = -2 \text{ or } n = 7.$

Since, no. of term cannot be negative.

∴ n = 7.

Substituting value of n in equation (1), we get :

⇒ a = 6 - 2n = 6 - 2 × 7 = 6 - 14 = -8.

Hence, a = -8 and n = 7.