Question

The product of three consecutive therms in a geometric progression is $216$ and sum their products in pairs is $156$. Find the three terms.

Answer 1

Let $a,ar$ and $a{r}^2$ be three consecutive terms in geometric progression.

As given, $\left(a\right)\left(ar\right)\left(a{r}^2\right)=216$ .......... $\left(1\right)$

and $\left(a\right)\left(ar\right)+\left(a\right)\left(a{r}^2\right)+\left(ar\right)\left(a{r}^2\right)=156$ ............ $\left(2\right)$

From $\left(1\right)$ we get, $a^3{r}^3=216$

Take cube root on both the sides

$⟹$ $ar=6$

$⟹$ $a=\frac{6}{r}$

Substitute the value of $a$ in equation (2) we get

$\left(\frac{6}{r}\right)(\frac{6}{r}\times r)+\left(\frac{6}{r}\right)(\frac{6}{r}\times r^2)+(\frac{6}{r}\times r)(\frac{6}{r}\times r^2)=156$

$⟹$ $\frac{36}{r}+36+36r=156$

$⟹$ $36+36r^2=120r$

$⟹$ $3r^2-10r+3=0$

$⟹$ $3r^2-9r-r+3=0$

$⟹$ $3r(r-3)-(r-3)=0$

$⟹$ $(r-3)(3r-1)=0$

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

(i) If $r=3$ then $a=2$

So, three consecutive terms would be $2,6,18$.

(ii) If $r=\frac{1}{3}$ then $a=18$

So, we get $18,6,2$.

$\therefore$ The three consecutive terms are $2,6,18$.