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

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#### Solution

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)

⇒ 3*a* = 24

∴ *a* = 8

(*a* – *d*) *a* (*a* + *d*) = 440 … (2)

⇒ (8 – *d*) (8) (8 + *d*) = 440

⇒ (8 – *d*) (8 + *d*) = 55

⇒ 64 – *d*^{2} = 55

⇒ *d*^{2} = 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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