Question

Find ${12}^{\mathrm{th}}$ term of a GP, whose $8^{\mathrm{th}}$ term is $192$ and common ratio is $2$ .

Answer 1

Step 1. Find the first term of GP:

We know that $n^{th}$ term of a G.P. is given as
$a_n=a{r}^{n-1}$
Where $a_n=n^{th}$ term of the G.P., $n$ is the number of terms, $a$ is the first term, and $r$ is the common ratio.

Here, common ratio, $r=2$.

$8^{th}$ term $a_8=192$
$\begin{array}{cc}\Rightarrow & \begin{array}{rcl}a{\left(2\right)}^{8-1}& =& 192\end{array}\\ \Rightarrow & \begin{array}{rcl}a{\left(2\right)}^7& =& 192\end{array}\\ \Rightarrow & \begin{array}{rcl}a\times 128& =& 192\end{array}\\ \Rightarrow & \begin{array}{rcl}a& =& \frac{192}{128}\end{array}\\ \Rightarrow & \begin{array}{rcl}a& =& \frac{3}{2}\end{array}\end{array}$

Therefore, the first term of GP is $a=\frac{3}{2}$.

Step 2: Find ${12}^{\mathrm{th}}$ term of a GP:

$\begin{array}{rcl}& \Rightarrow & a_{12}=a{r}^{12-1}\\ & =& \frac{3}{2}\times {\left(2\right)}^{11}\\ & =& \frac{3}{2}\times 2048\\ & =& 3\times 1024\\ & =& 3072\end{array}$

Hence, the value of ${12}^{th}$ term of the G.P. is $3072$.