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Question

The sum of three numbers in AP is 21 and their product is 231. Find the numbers.

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Solution

Let the required numbers be (ad),a,(a+d).
Then,
ad+a+a+d=21
3a=21
a=7
Also,
(ad)a(a+d)=231
a(a2d2)=231
7(49d2)=231
7d2=112
d2=16
d=±4
Hence, the required numbers are (3,7,11) or (11,7,3).

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