Question

Find three numbers in A.P. whose sum is 9 and sum of whose cubes in 99.

Answer 1

Given:

Let the three numbers in A.P. be a − d, a, a + d

The sum of the three numbers is 9

(a − d) + a + (a + d) = 9

3a = 9

⇒ a = 3

So the numbers are 3 − d, 3, 3 + d

The sum of their cubes is 99

(3 − d)³ + 3³ + (3 + d)³ = 99

(27 − 27d + 9d² − d³) + 27 + (27 + 27d + 9d² + d³) = 99

81 + 18d² = 99

18d² = 18

⇒ d² = 1

⇒ d = 1

Final numbers = 3 − 1, 3, 3 + 1

= 2, 3, 4

The three numbers in A.P. are 2, 3, and 4.