Question

If the product of three numbers in GP is 216 and the sum of their products in pairs is 156, find the numbers.

Answer 1

Let the required numbers be $\frac{a}{r}$, a and ar. Then,

$\frac{a}{r}\times a\times ar=216\Rightarrow a^3=216=6^3\Rightarrow a=6$

And, $\frac{a}{r}\times a+a\times ar+\frac{a}{r}\times ar=156$

$\Rightarrow a^2(\frac{1}{r}+r+1)=156\Rightarrow \left(6^2\right)(1+r+r^2)=156r[\because a=6]$

$\Rightarrow 36(r^2+r+1)=156r\Rightarrow 3(r^2+r+1)=13r$

$\Rightarrow 3r^2-10r+3=0\Rightarrow (3r-1)(r-3)=0\Rightarrow r=\frac{1}{3}$ or $r=3$

So, the required numbers are 18, 6, 2 or 2, 6, 18.