If the sum of first n terms of an A.P. is 3n² + 2n, find its r^th term.

By formula,

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

Given,

S_n = 3n² + 2n

Equating values we get :

$\Rightarrow 3n^2 + 2n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 3n^2 + 2n = \dfrac{n}{2}[a + a + (n - 1)d] \\[1em] \Rightarrow n(3n + 2) = \dfrac{n}{2}[a + a$n] \\[1em] \Rightarrow 3n + 2 = \dfrac{a + an}{2} \\[1em] \Rightarrow 2(3n + 2) = a + an \\[1em] \Rightarrow 6n + 4 = a + an \text{ ………(1)}

As,

S_n = 3n² + 2n

S₁ = 3(1)² + 2(1) = 3 + 2 = 5.

∴ a = 5.

Substituting value in (1), we get :

⇒ 6n + 4 = 5 + a_n

⇒ a_n = 6n - 1

⇒ a_r = 6r - 1

Hence, r_th term = 6r - 1.