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

Let the three numbersin A.P. be *a* – *d*, *a*, and *a* + *d*.

According to the giveninformation,

(*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, thenumbers are 11, 8, and 5.

Thus, the three numbersare 5, 8, and 11.

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