Question

Four numbers are in A.P. If their sum is 20 and the sum of their square is 120, then find the middle terms

Answer 1

Let the numbers are a – 3d, a – d, a + d, a + 3d

given a – 3d + a – d + a + d + a + 3d = 20

⇒ 4a = 20 ⇒ a = 5

and `(a – 3d)^2 + (a – d)2 + (a + d)^2 + (a + 3d)^2 = 120`

`4a^2 + 20 d^2 = 120`

`4 × 5^2 + 20 d^2 = 120`

d² = 1 ⇒ d = ±1

Hence numbers are 2, 4, 6, 8