The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[2 \times 20 + (7 - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[40 + 6d] \\[1em] \Rightarrow 2100 = 7(20 + 3d) \\[1em] \Rightarrow 300 = 20 + 3d \\[1em] \Rightarrow 3d = 280 \\[1em] \Rightarrow d = \dfrac{280}{3}.$

We know that,

a_n = a + (n - 1)d

⇒ 31^st term = a₃₁

= a + (31 - 1)d

= $20 + 30 \times \dfrac{280}{3}$

= 20 + 2800

= 2820.

Hence, 31^st term of A.P. = 2820.