Question

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

Answer 1

Let the four numbers in A.P be (a – 3d), (a – d), (a + d) and (a + 3d).

Then, (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20 

`\implies` 4a = 20

`\implies` a = 5

It is given that 

(a – 3d)² + (a – d)² + (a + d)² + (a + 3d)² = 120

`\implies` a² + 9d² – 6ad + a² + d² – 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 120

`\implies` 4a² + 20d² = 120

`\implies` a² + 5d² = 30

`\implies` 5² + 5d² = 30

`\implies` 25 + 5d² = 30

`\implies` 5d² = 5

`\implies` d² = 1

`\implies` d = ±1

When a = 5, d = 1

a – 3d = 5 – 3(1) = 2

a – d = 5 – 1 = 4

a + d = 5 + 1 = 6

a + 3d = 5 + 3(1) = 8

When a = 5, d = –1

a – 3d = 5 – 3(–1) = 8

a – d = 5 – (–1) = 6

a + d = 5 + (–1) = 4

a + 3d = 5 + 3(–1) = 2

Thus, the four parts are (2, 4, 6, 8) or (8, 6, 4, 2).