Question

The $4^{th}$ and $7^{th}$ terms of a $GP$ are $\frac{1}{27}$ and $\frac{1}{729}$ respectively. Find the sum of $m$ terms of the $GP$.

Answer 1

$a_4=\frac{1}{27}{a}_7=\frac{1}{729}$

$\Rightarrow a{r}^3=\frac{1}{27}.....\left(1\right)\Rightarrow a{r}^6=\frac{1}{729}.....\left(2\right)$

dividing $\left(2\right)$ in $\left(1\right)$

$\frac{a{r}^6}{a{r}^3}=\frac{\frac{1}{729}}{\frac{1}{27}}$

$\Rightarrow r^3=\frac{27}{729}$

$\Rightarrow r^3=\frac{1}{27}$

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

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

$a{\left(\frac{1}{3}\right)}^3=\frac{1}{27}$

$a\times \frac{1}{27}=\frac{1}{27}$

$\Rightarrow a=1$

So ,

$S_m=\frac{a\left(r^m-1\right)}{r-1}$

$=\frac{a{\left(\frac{1}{3}\right)}^3-1}{\frac{1}{3}-1}$

$=\frac{\frac{1}{3^m}-1}{-\frac{2}{3}}$

$=-\frac{3}{2}\left(\frac{1-3^m}{3^m}\right)$