Question

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

• A. $3,4,5$
• B. $2,4,6$
• C. $2,5,8$
• D. $3,6,9$

Answer 1

Answer: (B)

$2,4,6$

Let the numbers be $a,a+d,a+2d$

Given that $S_3=12$ that is:

$a+(a+d)+(a+2d)=12\Rightarrow a+a+d+a+2d=12\Rightarrow 3a+3d=12\Rightarrow 3(a+d)=12\Rightarrow a+d=\frac{12}{3}\Rightarrow a+d=4\Rightarrow a=4-d....\left(1\right)$

And it is given that the sum of their cubes is $288$, therefore, we have:

$a^3+(a+d{)}^3+(a+2d{)}^3=288\Rightarrow a^3+(a^3+d^3+3a^{2}d+3a{d}^2)+[a^3+(2{d)}^3+(3{\times a}^2\times 2d)+(3\times a\times (2d{)}^2]=288(\because (x+y{)}^3=x^3+y^3+3x^{2}y+3x{y}^2)\Rightarrow a^3+(a^3+d^3+3a^{2}d+3a{d}^2)+(a^3+8d^3+6a^{2}d+12a{d}^2)=288\Rightarrow a^3+a^3+a^3+d^3+8d^3+3a^{2}d+6a^{2}d+3a{d}^2+12a{d}^2=288\Rightarrow 3a^3+9d^3+9a^{2}d+15a{d}^2=288\Rightarrow 3{(4-d)}^3+9d^3+9{(4-d)}^{2}d+15(4-d)d^2=288\left(Usingequation1\right)\Rightarrow 3[4^3-d^3-(3{\times 4}^2\times d)+(3\times 4\times d^2\left)\right]+9d^3+9[4^2+d^2-(2\times 4\times d\left)\right]d+(60-15d)d^2=288(\because (x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2),(x-y{)}^2=x^2+y^2-2xy)\Rightarrow 3(64-d^3-48d+12d^2)+9d^3+9(16+d^2-8d)d+(60-15d)d^2=288\Rightarrow 192-3d^3-144d+36d^2+9d^3+144d+9d^3-72d^2+60d^2-15d^3=288\Rightarrow 192-3d^3+9d^3+9d^3-15d^3+36d^2-72d^2+60d^2+144d-144d=288\Rightarrow 24d^2=288-192\Rightarrow 24d^2=96\Rightarrow d^2=\frac{96}{24}\Rightarrow d^2=4\Rightarrow d=\pm \sqrt{4}\Rightarrow d=\pm 2$

For $d=2,a=4–d=4–2=2$

Then the numbers will be $2,4$ and $6$.

For $d=-2,a=4-(-2)=4+2=6$

Then the numbers will be $6,4$ and $2$.

Hence, the required numbers are $2,4$ and $6$.