Question

The fourth term of an Arithmetic progression is 10 & eleventh term of it exceeds three times the fourth term by 1. Find the sum of the first 20 terms of the progression,​

Answer 1

Answer:

The 4th term is a + 3d = 10 … (1)

The 11th term is (a + 10d) - 3(a + 3d) = 1, or

a + 10d - 3a - 9d = 1, or

-2a + d = 1 … (2)

From (1) a = 10 - 3d. Put that in (2) to get

-2(10 - 3d) + d = 1, or

-20 + 6d + d = 1, or

7d = 21, or

d = 3.

From (2) d-1 = 2a, 3–1 = 2 = 2a, or a = 1

Sn = (n/2)[2a + (n-1)d], or

S20 = (20/2)[2*1 + (20–1)*3]

= 10[2 + 19*3]

= 10(2+57)

= 10*59

= 590

Answer = 590.