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.


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