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

Let three numbers in A.P. be (a - d), a, (a + d).

Sum = 24

∴ a - d + a + a + d = 24

⇒ 3a = 24

⇒ a = 8.

Product = 440

⇒ (a - d)(a)(a + d) = 440

⇒ (8 - d)(8)(8 + d) = 440

⇒ (8 - d)(8 + d) = $\dfrac{440}{8}$

⇒ 64 - d² = 55

⇒ d² = 64 - 55 = 9

⇒ d = ± 3

Let d = 3,

A.P. = (8 - 3), 8, (8 + 3) = 5, 8, 11.

Let d = -3,

A.P. = (8 - (-3)), 8, (8 + (-3)) = 11, 8, 5.

Hence, A.P. = 5, 8, 11 or 11, 8, 5.