Question

Which term of the G.P$=2,2\sqrt{2},4,....$ ________is $128$.

• A. $13$
• B. $12$
• C. $11$
• D. $10$

Answer 1

The correct option is A $13$

We know that $a_n=a{r}^{n-1}$

where,

$a_n=n^{th}termofGP$

$n$ is the number of terms

$a$ is the first term

$r$ is the common ratio.

Here,

First term $a=2$

common ratio

$\begin{array}{c}r=\frac{2\sqrt{2}}{2}\\ =\sqrt{2}\end{array}$

$n^{th}$ term $=128$

putting values

$n^{th}term=a{r}^{n-1}$

$\begin{array}{c}128=\left(2\right){\left(\sqrt{2}\right)}^{n-1}\\ \frac{128}{2}={\left(\sqrt{2}\right)}^{n-1}\\ 64={\left(\sqrt{2}\right)}^{n-1}\\ 2^6={\left(\sqrt{2}\right)}^{n-1}\end{array}$

Now, squaring both sides

$\begin{array}{c}{\left(2^6\right)}^2={\left[{\left(\sqrt{2}\right)}^{n-1}\right]}^2\\ 2^{6\times 2}={\left[{\left(\sqrt{2}\right)}^2\right]}^{n-1}\\ 2^{12}={\left[2\right]}^{n-1}\end{array}$

And, comparing powers we get

$\begin{array}{c}12=n-1\\ 12+1=n\\ 13=n\\ n=13\end{array}$

Therefore, ${13}^{th}$ term of GP is $128$

Hence, Option $A$ is the correct answer.