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

Let three consecutive terms be (a - d), a, (a + d)

S = 21

⇒ a - d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7.

Given, sum of their squares is 165

⇒ (a - d)² + a² + (a + d)² = 165

⇒ (7 - d)² + 7² + (7 + d)² = 165

⇒ 49 + d² - 14d + 49 + 49 + d² + 14d = 165

⇒ 147 + 2d² = 165

⇒ 2d² = 165 - 47 = 18

⇒ d² = 9

⇒ d = ± 3

Let d = 3,

A.P. = (7 - 3), 7, (7 + 3) = 4, 7, 10.

Let d = -3,

A.P. = (7 - (-3)), 7, (7 + (-3)) = 10, 7, 4.

Hence, the three consecutive terms of A.P. = 4, 7, 10 or 10, 7, 4.