In an A.P., given a = 2, d = 8, S_n = 90, find n and a_n.

By formula,

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

Given,

S_n = 90

$\Rightarrow \dfrac{n}{2}[2 \times 2 + (n - 1) \times 8] = 90 \\[1em] \Rightarrow n[4 + 8n - 8] = 180 \\[1em] \Rightarrow n[8n - 4] = 180 \\[1em] \Rightarrow 8n^2 - 4n = 180 \\[1em] \Rightarrow 4(2n^2 - n) = 180 \\[1em] \Rightarrow 2n^2 - n = \dfrac{180}{4} \\[1em] \Rightarrow 2n^2 - n = 45 \\[1em] \Rightarrow 2n^2 - n - 45 = 0 \\[1em] \Rightarrow 2n^2 - 10n + 9n - 45 = 0 \\[1em] \Rightarrow 2n(n - 5) + 9(n - 5) = 0 \\[1em] \Rightarrow (2n + 9)(n - 5) = 0 \\[1em] \Rightarrow 2n + 9 = 0 \text{ or } n - 5 = 0 \\[1em] \Rightarrow 2n = -9 \text{ or } n = 5 \\[1em] \Rightarrow n = -\dfrac{9}{2} \text{ or } n = 5.$

Since, no. of terms cannot be negative.

∴ n = 5.

By formula,

a_n = a + (n - 1)d

a₅ = 2 + (5 - 1) × 8

= 2 + 4 × 8

= 2 + 32 = 34.

Hence, n = 5 and a_n = 34.