Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Let numbers be a - 3d, a - d, a + d, a + 3d.

Given, sum = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 5.

Given, sum of squares is 120.

⇒ (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120

⇒ (5 - 3d)² + (5 - d)² + (5 + d)² + (5 + 3d)² = 120

⇒ 25 + 9d² - 30d + 25 + d² - 10d + 25 + d² + 10d + 25 + 9d² + 30d = 120

⇒ 100 + 20d² = 120

⇒ 20d² = 20

⇒ d² = 1.

⇒ d = ±1

Let d = 1,

Numbers = (5 - 3(1)), (5 - 1), (5 + 1), (5 + 3(1))

= 2, 4, 6, 8.

Let d = -1,

Numbers = (5 - 3(-1)), (5 - (-1)), (5 + (-1)), (5 + 3(-1))

= 8, 6, 4, 2.

Hence, numbers are 2, 4, 6, 8 or 8, 6, 4, 2.