Question

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.

Answer 1

Let the three numbers be $\frac{a}{r},a,ar$

$\Rightarrow \frac{a}{r}\times a\times ar=1000$

$\Rightarrow a^3=1000$

$\Rightarrow a=10$

$2(a+6)=(ar+7)+\frac{a}{r}$ where $a=10$

$\Rightarrow 32r=10r^2+7r+10$

$\Rightarrow 10r^2+7r-32r+10=0$

$\Rightarrow 10r^2-25r+10=0$

$\Rightarrow 2r^2-5r+2=0$

$\Rightarrow 2r^2-4r-r+2=0$

$\Rightarrow 2r(r-2)-1(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r=2,\frac{1}{2}$

$\therefore a=10,r=2\Rightarrow$ terms are $\frac{10}{2},10,10\times 2$ or $5,10,20$

$\therefore a=10,r=\frac{1}{2}\Rightarrow$ terms are $\frac{10}{\frac{1}{2}},10,10\times \frac{1}{2}$ or $20,10,5$