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Question

The sum of three numbers in A.P. is 12, and the sum of their
cubes is 408; find them.

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Solution

Let 3 numbers in A.P. be (ad),a,(a+d)
Sum of 3 numbers =12
ad+a+a+d=12
3a=12
a=4
sum of their cubes =408
(ad)3+a3+(a+d)3=408
a3d33a2d+3ad2+a3+a3+d3+3a2d+3ad2=408
3a3+6ad2=408
a3+2ad2=136
(4)3+2(4)d2=136
8d2=13664
8d2=72
d2=9
d=±3
d=3, series 43,4,4+3
1,4,7
d=3, series 4(),4,43
7,4,1
3 numbers are 1,4 and 7.

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