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

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

According to the giveninformation,

(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 – d²= 55

⇒ d²= 64 – 55 = 9

⇒ d = ±3

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

Thus, the three numbersare 5, 8, and 11.