Question

Let $a_n$ denote the $n^{\text{th}}$ term of a G.P. If $a_1=3,a_n=96$ and sum of $n$ terms of the series is $189$, then the value of $n$ is

Answer 1

Let $r$ be the common ratio of the given G.P., so
$a_1=a{a}_n=96\Rightarrow a{r}^{n-1}=96\Rightarrow 3\cdot r^{n-1}=96\Rightarrow r^{n-1}=32\cdots \left(i\right)$

Now, $S_n=189$
$\Rightarrow a\left(\frac{r^n-1}{r-1}\right)=189\Rightarrow 3\left(\frac{32r-1}{r-1}\right)=189\Rightarrow \frac{32r-1}{r-1}=63\Rightarrow 31r=62\Rightarrow r=2$
Using equation $\left(1\right)$, we get
$2^{n-1}=32\therefore n=6$