Question

Find three numbers are A. P. whose sum is 24 and whose product is 440.

Answer 1

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

According to the given information,

(a – d) + (a) + (a + d) = 24 … (1)

⇒ 3a = 24

∴ a = 8

(a – d) a (a + d) = 440 … (2)

⇒ (8 – d) (8) (8 + d) = 440

⇒ (8 – d) (8 + d) = 55

⇒ 64 – $d^2$ = 55

⇒ $d^2$ = 64 – 55 = 9

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.