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Question

Find three consecutive terms in an A.P. whose sum is 3 and the product of their cubes is 512.

A
4,1,2
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B
4,1,2
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C
4,2,3
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D
None of these
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Solution

The correct option is A 4,1,2

Let the numbers be (ad),a,(a+d). Then,

Sum=3(ad)+a+(a+d)=33a=3a=1

Product of their cubes=512(ad)3×a3×(a+d)3=512

(a2d2)3×a3=512

(1d2)3=512 (a=1)

(1d2)=8

d2=9d=±3

If d=3, then the numbers are 4,1,2. If d=3, then the numbers are 2,1,4.


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