Question

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

• A. $2,4,6$ or $6,4,2$
• B. $3.4,6$ or $8,4,3$
• C. $7,8,9$ or $10,9,8$
• D. $0,4,6$ or $7,4,2$

Answer 1

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

Consider the numbers are $a,a+d,a+2d$

Given that $S_3=12$

$\Rightarrow a+a+d+a+2d=12$

$3(a+d)=12$

$a+d=4$

$a=4-d$------------(1)

And sum of their cubes is $288.$

$(a{)}^3+(a+d{)}^3+(a+2d{)}^3=288$

$a^3+a^3+d^3+3a^{2}d+3a{d}^2+a^3+8d^3+6a^{2}d+12a{d}^2=288$

$3a^3+9d^3+9a^{2}d+15a{d}^2=288$

$3(4-d{)}^3+9d^3+9(4-d{)}^{2}d+15(4-d)d^2=288$ [using equation 1]

$3(64–d^3-48d+12d^2)+9d^3+9(16+d^2-8d)d+(60-15d)d^2=288$

$192-3d^3-144d+36d^2+9d^3+144d+9d^3-72d^2+60d^2-15d^3=288$

$24d^2=288-192$

$24d^2=96$

$d^2=96/24$

$d^2=4$

$d=\pm 2$

For $d=2,a=4–d=4–2=2$

The numbers will be $2,4and6.$

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

The numbers will be $6,4and2.$

Hence, the required numbers are $(2,4,6)or(6,4,2)$