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Question

# The sum of three number in A.P is 12 and the sum of their cubes is 288. Find the numbers.

A
3,4,5
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B
2,4,6
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C
2,5,8
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D
3,6,9
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Solution

## The correct option is C 2,4,6Let the numbers be a,a+d,a+2dGiven that S3=12 that is:a+(a+d)+(a+2d)=12⇒a+a+d+a+2d=12⇒3a+3d=12⇒3(a+d)=12⇒a+d=123⇒a+d=4⇒a=4−d....(1)And it is given that the sum of their cubes is 288, therefore, we have:a3+(a+d)3+(a+2d)3=288⇒a3+(a3+d3+3a2d+3ad2)+[a3+(2d)3+(3×a2×2d)+(3×a×(2d)2]=288(∵(x+y)3=x3+y3+3x2y+3xy2)⇒a3+(a3+d3+3a2d+3ad2)+(a3+8d3+6a2d+12ad2)=288⇒a3+a3+a3+d3+8d3+3a2d+6a2d+3ad2+12ad2=288⇒3a3+9d3+9a2d+15ad2=288⇒3(4−d)3+9d3+9(4−d)2d+15(4−d)d2=288(Usingequation1)⇒3[43−d3−(3×42×d)+(3×4×d2)]+9d3+9[42+d2−(2×4×d)]d+(60−15d)d2=288(∵(x−y)3=x3−y3−3x2y+3xy2),(x−y)2=x2+y2−2xy)⇒3(64−d3−48d+12d2)+9d3+9(16+d2−8d)d+(60−15d)d2=288⇒192−3d3−144d+36d2+9d3+144d+9d3−72d2+60d2−15d3=288⇒192−3d3+9d3+9d3−15d3+36d2−72d2+60d2+144d−144d=288⇒24d2=288−192⇒24d2=96⇒d2=9624⇒d2=4⇒d=±√4⇒d=±2For d=2,a=4–d=4–2=2Then the numbers will be 2,4 and 6.For d=−2,a=4−(−2)=4+2=6Then the numbers will be 6,4 and 2.Hence, the required numbers are 2,4 and 6.

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