Question

The sum in of three numbers AP is $12$ and the sum of their cubes is $288$. Find the numbers.

• A. $2,4,6or6,4,2$
• B. $4,6,8or8,6,4$
• C. $10,12,14or14,12,10$
• D. $-2,-4,-6or-6,-4,-2$

Answer 1

The correct option is A $2,4,6or6,4,2$

Let the numbers be $(a-d),a,(a+d)$. Then,
$Sum=12\Rightarrow (a-d)+a+(a+d)=12\Rightarrow 3a=12\Rightarrow a=4$
$Sumoftheircubes=288\Rightarrow (a-d{)}^3+a^3+(a+d{)}^3=288$
$\Rightarrow (a^3-3a^{2}d+3a{d}^2-d^3)+a^3+(a^3+3a^{2}d+3a{d}^2+d^3)=288$
$\Rightarrow 3a^3+6a{d}^2=288$
$\Rightarrow 3\times 64+6\times 4\times d^2=288$ ($\because a=4$)
$\Rightarrow d^2=4\Rightarrow d=\pm 2$
If d=2, then the numbers are $2,4,6$. If d=-2, then the numbers are $6,4,2$
Thus, the numbers are $2,4,6$ or $6,4,2.$