Question

If the product of three consecutive term in $G.P$ is $216$ and sum of their products in pairs is $156$, find them.

Answer 1

Let $\frac{a}{r},a$ ar be three consecutive term in G.P.

with common ratio 'r'

then according to question

$\frac{a}{r}\times a\times ar=216$ ------ (1)

$a^3=216$

[a=6]

and

$\frac{a^2}{r}+a^{2}r+a^2=156$

$a^2[\frac{1}{r}+r+1]=156$

$\frac{r^2+r+1}{r}=\frac{156}{36}$

$\Rightarrow$ $3r^2+3r+3=13r$

$\Rightarrow$ $3r^2-10r+3=0$

$3r^2-9r-r+3=0$

(3r-1)(r-3)=0

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

$\therefore$ three numbers are 18,6,2 or 2,6,18.