Question

Find the sum of first ${12}^{th}$ term of the GP, whose $8^{th}$ term is $192$ and the 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, the common ratio, $r=2,8^{th}$ term, $a_8=192$

We have to find the value of $a_{12}$.

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

Thus, the first term of GP is $\frac{3}{2}$.

Step 2. Find the sum of first ${12}^{th}$ term of the GP:

We know that the sum of $n$ terms of the G.P. is
$S_n=\frac{a\left(r^n-1\right)}{r-1}\left[\because r>1\right]$
And $S_n=\frac{a\left(1-r^n\right)}{1-r}\left[\because r<1\right]$

Where $a=$ first term, $r=$ common ratio, and $n=$ number of terms

Here, $a=\frac{3}{2}$

$\therefore r=2>1$

Now, the sum of ${12}^{th}$ terms is given as

$\begin{array}{rcl}& \Rightarrow & S_{12}=\frac{\frac{3}{2}[{\left(2\right)}^{12}-1}{2-1}\\ & =& \frac{3\left(4096-1\right)}{2\times 1}\\ & =& \frac{3\times 4095}{2}\\ & =& \frac{12285}{2}\\ & =& 6142.5\end{array}$

Hence, the sum of first ${12}^{th}$term of the GP is $6142.5.$