Question

The sum of the first three terms of an increasing G.P. is $13$ and their product is $27$. The sum of the first five terms of the G.P. is:

Answer 1

Let the terms be $\frac{a}{r},a,ar$

Given $\frac{a}{r}+a+ar=13$ ………….$\left(1\right)$

$\left(\frac{a}{r}\right)\left(a\right)\left(ar\right)=27$

$a^3=27$,

$a=3$

Now $\left(1\right)$

$\Rightarrow \frac{3}{r}+3+3r=13$

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

$3+3r^2=10r$

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

$3r^2-9r-8+3=0$

$3r(r-3)-(r-3)=0$

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

Given increasing g.p $\therefore r=3$

$\therefore$ Sum of five terms $=\frac{a}{r}+a+ar+a{r}^2+a{r}^3$.

We know $\frac{a}{r}+a+ar=13$

$a=3,r=3$

$a{r}^2+a{r}^3=a{r}^2(1+r)$

$=3\left(3^2\right)(1+3)$

$=27\times 4$

$=108$

\Sum of five terms $=108+13=121$.