Question

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

Answer 1

Let the required terms be (a – d), a and (a + d). 

Then (a – d) + a + (a + d) = 21

⇒ 3a = 21 

⇒ a = 7 

Also, (a – d)² + a² + (a + d)² = 165

⇒ 3a² + 2d² = 165

⇒ (3 × 49 + 2d²) = 165

⇒ 2d² = 165 – 147 = 18

⇒ d² = 9

⇒ d = ±3

Thus, a = 7 and d = ±3

Hence, the required terms are (4, 7, 10) or (10, 7, 4).