Question

In a G.P, it is being given that $T_1=3$, $T_n=96$ and $S_n=189$. Then the value of $n$ is
(where $T_n$ and $S_n$ denote the $n^{th}$ term and sum upto $n^{th}$ term repectively)

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

Given
$a=3,T_n=a{r}^{n-1}=96S_n=189$
Let the common ratio of the given G.P. be $r$.
Then,
$S_n=\frac{a(r^n-1)}{(r-1)}\Rightarrow 189=\frac{a(r^n-1)}{(r-1)}\Rightarrow \frac{(a{r}^n-a)}{(r-1)}=189\Rightarrow \frac{(r{T}_n-a)}{(r-1)}=189$
$\Rightarrow \frac{(96r-3)}{(r-1)}=189$
$\Rightarrow (96r-3)=(189r-189)$
$\Rightarrow 93r=186\Rightarrow r=2$
Now,
$T_n=a{r}^{n-1}$
$\Rightarrow 3\times 2^{n-1}=96$
$\Rightarrow 2^{n-1}=32=2^5$
$\Rightarrow n-1=5\therefore n=6$