Question

if products of three terms of a GP is 216 and sum of their products taken in pairs is 156, then find the numbers.

Answer 1

Let the nos, be

$\begin{array}{c}\frac{\alpha }{r},\alpha ,\alpha r\\ \frac{\alpha }{r}\times \alpha \times \alpha r=216\\ {\alpha }^3={\left(6\right)}^3\\ \Rightarrow \alpha =6\frac{\alpha }{2}=\frac{6}{3}=2\alpha r=18\left(r=3\right)\end{array}$

Hence nos are $2,6$ and$18$.

$\frac{{\alpha }^2}{r}\times {\alpha }^2\times {\alpha }^{2}r=156$

$36\left[r+\frac{1}{r}+1\right]=156$

$r+\frac{1}{r}=\frac{13}{3}-1$

$r+\frac{1}{r}=\frac{10}{3}$

$\frac{r^2+1}{r}=\frac{10}{3}$

$3r^3-10r+3=0$

$3r^3-9r-r+3=0$

$3r\left(r-3\right)-1\left(r-3\right)=0$

$r=\frac{1}{3}r=3$