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.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 21

∴ (a - d) + a + (a + d) = 21

⇒ 3a = 21

⇒ a = 7

Now, sum of the squares of the terms = 165

⇒ (a - d)² + a² + (a + d)² = 165

⇒ a² + d² - 2ad + a² + a² + d² + 2ad = 165

⇒ 3a² + 2d² + a = 165

Put the value of a = 7, we get,

3(49) + 2d² = 165

⇒ 2d² = 165 - 147

⇒ 2d² = 18

⇒ d² = 9

⇒ d = 3

∴ If d = 3, then the numbers are 4, 7, 10.

If d = - 3, then the numbers are 10, 7, 4.