Question

If the $4^{th}$ and $9^{th}$ term of a G.P. are $54$ and $13122$ respectively, then the ${15}^{th}$ term is

• A. $2\times 3^{15}$
• B. $2\times 2^{14}$
• C. $3^{14}$
• D. $2\times 3^{14}$

Answer 1

The correct option is D $2\times 3^{14}$

Let $a$ be the first term and $r$ be the common ratio of the given G.P.
Then, $T_4=54$
$\Rightarrow a{r}^{(4-1)}=54\Rightarrow a{r}^3=54\dots \left(1\right)$
And, $T_9=13122$
$\Rightarrow a{r}^{(9-1)}=13122\Rightarrow a{r}^8=13122\dots \left(2\right)$

On dividing $\left(2\right)$ by $\left(1\right),$ we get
$\frac{a{r}^8}{a{r}^3}=\frac{13122}{54}$
$\Rightarrow r^5=243=3^5\Rightarrow r=3$

Putting $r=3$ in $\left(1\right),$ we get
$a\times 3^3=54$
$\Rightarrow 27a=54\Rightarrow a=2$
Thus, $a=2$ and $r=3$
So, the required G.P. is $2,6,18,54,\dots$
The general term of the G.P. is given by
$T_n=a{r}^{(n-1)}=2\times 3^{(n-1)}$
$\Rightarrow T_{15}=2\times 3^{14}$