Question

The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers. 

Answer 1

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

Then, (a – d) + a + (a + d) = 15

`\implies` 3a = 15

`\implies` a = 5

It is given that 

(a – d)² + (a + d)² = 58

`\implies` a² + d² – 2ad + a² + d² + 2ad = 58

`\implies` 2a² + 2d² = 58

`\implies` 2(a² + d²) = 58

`\implies` a² + d² = 29

`\implies` 5² + d² = 29

`\implies` 25 + d² = 29

`\implies` d² = 4

`\implies` d = ±2

When a = 5 and d = 2

a – d = 5 – 2 = 3

a = 5

a + d = 5 + 2 = 7

When a = 5 and d = –2

a – d = 5 – (–2) = 7

a = 5

a + d = 5 + (–2) = 3

Thus, the three numbers in A.P. are (3, 5, 7) or (7, 5, 3)