The sum of first 14 terms of an A.P. is 1050 and its 14^th term is 140. Find the 20^th term.

Given,

⇒ a₁₄ = 140

⇒ a + (14 - 1)d = 140

⇒ a + 13d = 140

⇒ a = 140 - 13d …….(i)

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 1050 = \dfrac{14}{2}[2 \times (140 - 13d) + (14 - 1)d] \\[1em] \Rightarrow 1050 = 7[280 - 26d + 13d] \\[1em] \Rightarrow 1050 = 7[280 - 13d] \\[1em] \Rightarrow 1050 = 1960 - 91d \\[1em] \Rightarrow 91d = 1960 - 1050 \\[1em] \Rightarrow 91d = 910 \\[1em] \Rightarrow d = 10.$

Substituting value of d in (i) we get,

⇒ a = 140 - 13(10) = 140 - 130 = 10.

a₂₀ = a + 19d = 10 + 19(10) = 200.

Hence, 20^th term = 200.