Question

Find four numbers in ap whose sum is 12 and sum of whose squares is 116

Answer 1

Solution :

Let, the numbers in AP are

(a - 3d), (a - d), (a + d) & (a + 3d)

Given that,

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 12

→ 4a = 12

→ a = 3

Also,

(a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 116

→ 4a² + 20d² = 116

→ 4 (3²) + 20d² = 116

→ 20d² = 116 - 36

→ 20d² = 80

→ d² = 4

∴ d = ± 2

When a = 4 and d = + 2, the numbers are

4 - 3 (2), 4 - 2, 4 + 2, 4 + 3 (2)

i.e., - 2, 2, 6 and 10

When a = 4 and d = - 2, the numbers are

4 - 3 (-2), 4 - (-2), 4 + (-2) and 4 + 3 (-2)

i.e., 10, 6, 2 and - 2

Hence, the required four numbers are

10, 6, 2 and - 2