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 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 – d2 = 55
⇒ d2 = 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.
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