The sum of three numbers in AP is 21 and their product is 232 find the numbers.
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Let the AP is (a-d), a, (a+d)
(a-d)+a+(a+d) =21
3a=21
a=7
(a-d)×a×(a+d)=231
(a×a-d×d) ×a=231
(7×7-d×d)×7=231
(7×7-d×d)=231÷7
(7×7-d×d)=33
d×d=49-33
d×d=16
d=4
Hence, the A. P. is 3, 7, 11.
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