Four numbers are in arithmetic progression. If their sum is 20 and the sum of their squares is 120, then find the numbers.

Let the first term be a – 3d and common difference be 2d.
Thus arithmetic progression will be a - 3d, a - d, a + d, a+3d.

Given, Sum of terms = 20
⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20
⇒ a = 5 …(i)

Also,

Sum of their square = 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

If d = 1

First term, a – 3d = 5 – 3 = 2 and common difference, 2d = 2 and four terms will be 2, 4, 6, 8

If d = -1

First term, a – 3d = 5 –(– 3) = 8 and common difference, 2d = -2 and four terms will be 8, 6, 4 and 2.