Question

In an A.P., 15^th term exceeds the 8^th term by 21. If sum of first 10 terms is 55, then form the A.Р.

Answer 1

Given: a₁₅ – a₈ = 21

(a + 14d) – (a + 7d) = 21

7d = 21

⇒ d = 3

Given: S₁₀ = 55

`10/2 [2a + (10 - 1)d] = 55`

5[2a + 9(3)] = 55

2a + 27 = 11

2a = 11 – 27

2a = –16

⇒ a = –8

The A.P. is:

Term 1: a = –8

Term 2: a + d = –8 + 3 = –5

Term 3: a + 2d = –8 + 6 = –2

Term 4: a + 3d = –8 + 9 = 1

The A.P. is –8, –5, –2, 1, ...