Question

What is the sum of the first 20 terms of an A.P., if a=4 and 1 = 36?

(A) 40 (B) 200 (C) 400 (D) 800​

Answer 1

Given:

a = 4

d = 36

n = 20

Now,

$s_{20} = \frac{n}{2} [2a + (n - 1)d ]$

$= \frac{20}{2} [2(4) + (20 - 1)36 ]$

$= 10 [8 + (19)36 ]$

$= 10 [8 + 684 ]$

$= 10(692)$

$= 6920$