Question

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

Answer 1

Let the numbers be,

$a-d,a,a+d$

Given,

$a-d+a+a+d=12$

$3a=12$

$\Rightarrow a=4$

$(4-d{)}^3+4^3+(4+d{)}^3=288$

$24d^2+192=288$

$d^2=4$

$\Rightarrow d=\pm 2$

when $a=4,d=2$ the numbers are, $2,4,6$

when $a=4,d=-2$ the numbers are, $6,4,2$