Question

The product of three consecutive terms of a G.P. is $216$ . If $3$ is added to each of the first and the second of these terms, the three form an A.P. Find the sum of the original three terms.

Answer 1

Let the three terms in G.P. be $\frac{a}{r}, a$ and $a r$.
Product of three terms $=216$
$\Rightarrow \frac{a}{r} \cdot a \cdot a r=216 $
$\Rightarrow a^{3}=6^{3} $
$\Rightarrow a=6$
Three terms in A.P. are $\frac{6}{r}+3,6+3$ and $6 r$
i.e. $\frac{6}{r}+3,9$ and $6 r$
$\Rightarrow 18=\frac{6}{r}+3+6 r $
$\Rightarrow 2 r^{2}-5 r+2=0 $
$\Rightarrow(r-2)(2 r-1)$
$\Rightarrow r=2 $
or $ r=\frac{1}{2}$
The three terms in G.P. are $3,6,12$ or $12,6,3$.
$\Rightarrow $ Required sum $=3+6+12=21$