Question

Find the common ratio of the $G.P.$ whose first and last terms are $25$ and $\frac{1}{625}$, respectively and the sum of the $G.P.$ is $\frac{19531}{625}$.

Answer 1

Given that:-

First term of G.P. $\left(a\right)=25$

Last term of G.P. $\left(n\right)=\frac{1}{625}$

As we know that, $n^{th}$ term of a G.P. is given by-

$a_n=a{r}^{n-1}$

$\frac{1}{625}=25\times r^{n-1}$

$\Rightarrow r^n=\frac{r}{15625}.....\left(1\right)$

Sum of G.P. $=\frac{19531}{625}\left[\text{Given}\right]$

Sum of $n$ terms of G.P.-

$S_n=\frac{a(1-r^n)}{r-1}[\because r<1]$

$\therefore \frac{25(1-\frac{r}{15625})}{1-r}=\frac{19531}{625}$

$\Rightarrow 15625-r=19531-19531r$

$\Rightarrow 19530r=19531-15625$

$\Rightarrow r=\frac{3906}{19530}=\frac{1}{5}$

Hence the common ratio of the given G.P. is $\frac{1}{5}$.