# The product of $3$ numbers in $GP$ is $1000$. If $6$ and $7$ are added to the $2^{nd}$ and $3^{rd}$ terms respectively, the terms form an $AP.$ Find the $GP$.

$\begin{array}{1 1} 4,10,25 \\ 25,10,4 \\ 5,10,20 \\ -5,10,-20 \end{array}$

$Explanation : \;Let\;the\;3\;numbers\;be\;\frac{a}{r},a,ar.$
$\frac{a}{r}*a*ar=1000$
$a^3=1000$
$a=10$
$2\;(a+6)=(ar+7)+\frac{a}{r}\quad\;a=10$
$32r=10r^2+7r+10$
$10r^2-25r+10=0$
$r^2-5r+1=0$
$(r-2)(2r-1)=0\quad\;r=2,1/2$
$=5,10,20\quad\;or\;20,10,5.$