# Find 3 numbers in AP whose sum is 18 and the sum of its cube is 972.

$\begin{array}{1 1} -9,9,18 \\ 2,6 ,10 \\ 3,6,9 \\ 1,5, 9 \end{array}$

Explanation : Let the three numbers in AP be a-d,a,a+d
$a-d+a+a+d=18$
$3a=18$
$a=6$
$middle\; term is \;6$
$Given \;that\;(a-d)^3+a^3+(a+d)^3=972$
$a^3-3a^2d+3ad^2-d^3+a^3+a^3+3a^2d+3ad^2+d^3=972$
$3a^3+6ad^2=972$
a=6
$3*216+6*6d^2=972$
$d^2=\large\frac{972-648}{36}=9$
$d=3$
$a=6,\;d=3$
$numbers\;are\;a-d,a,a+d$
$=3,6,9.$
edited Jan 24, 2014