Question

The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.

Answer 1

Let the three consecutive terms in A.P. be a – d, a and a + d.

∴ (a – d) + a + (a + d) = 21

`=>` a = 7   ...(1) 

Also, (a – d)² + a² + (a + d)² = 165

`=>` a² + d² - 2ad + a² + a² + d² + 2ad = 165

`=>` 3a² + 2d² = 165

`=>` 3 × (7)² + 2d² = 165   ...[From (1)]

`=>` 3 × 49 + 2d² = 165 

`=>` 147 + 2d² = 165 

`=>` 2d² = 18

`=>` d² = 9

`=>` d = ±3 

When a = 7 and d = 3

Required terms = a – d, a and a + d

= 7 – 3, 7, 7 + 3

= 4, 7, 10 

When a = 7 and d = –3

Required terms = a – d, a and a + d

= 7 – (–3), 7, 7 + (–3)

= 10, 7, 4