If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

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

According to question -

(a – d) + (a) + (a + d) = 24 … (1)

⇒ 3a = 24

∴ a = 8

and,

(a – d) a (a + d) = 440 … (2)

⇒ (8 – d) (8) (8 + d) = 440

⇒ (8 – d) (8 + d) = 66

⇒ 64 – d2 = 66

⇒ d2 = 64 – 66 = 9

⇒ d = 3

Therefore,

when d = 3, the numbers are 6, 8, and 11 and

when d = –3, the numbers are 11, 8, and 6.

Thus, the three numbers are 6, 8, and 11.