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.