The 14th term of an A.P. is twice its 8th term. If its 6th term is −8, then find the sum of its first 20 terms.

Let the first term be a and common difference be d
a14 = a + (14-1)d = a + 13d
a8 = a + (8-1)d = a + 7d
a6 = a + (6-1)d = a + 5d
Given: a14 is twice of a8
∴ a + 13d = 2(a + 7d) ⇒ a + 13d = 2a + 14d ⇒ 2a + 14d - a - 13d = 0 ⇒ a + d = 0 ⇒ a = -d (1)
a6 = a + (6-1)d = a + 5d ∴ a + 5d = -8. (2)
Put value a = -d in equation (2) we get
-d + 5d = -8 ⇒ 4d = -8 ⇒ d = -2;
Substituting the value of d = -2; in equation (1) we get a = 2
Then S20 = 20/2[2a + (20-1)d] = 10[2(2) + (19)(-2)] = 10(4 - 38) = -340