Question

Find the side of a square whose area is $7^{th}$ term in $G.P$ of $2,\sqrt{8},\mathrm{4....}$

Answer 1

$2,\sqrt{8},4,.....$

For the given G.P.-

First term $\left(a\right)=2$

Common ratio $\left(r\right)=\frac{\sqrt{8}}{2}=\sqrt{2}$

As we know that, $n^{th}$ term in a G.P. is given as-

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

Therefore,

$a_7=2\times {\sqrt{2}}^{7-1}=16$

Now,

Area of square $=a_7=16\text{sq. units}$

Let $a$ be the side of square.

Therefore,

Area of square $=a^2$

$\Rightarrow a^2=16$

$\Rightarrow a=\sqrt{16}=4\text{unit}$