Question

Which term of the Geometric progression $2,2\sqrt{2},4,...$ is $64$?

Answer 1

Given series is $2,2\sqrt{2},4,...$
Here $a=2,T_n=64$
$\therefore r=\frac{2\sqrt{2}}{2}=\sqrt{2}$
We know $T_n=a.r^{n-1}$
$\Rightarrow 64=2.{\left(\sqrt{2}\right)}^{n-1}$
$\Rightarrow {\left(\sqrt{2}\right)}^{n-1}=32$
$\Rightarrow 2^{\frac{1}{2}(n-1)}=32$
$\Rightarrow 2^{\frac{n-1}{2}}=2^5$

As the powers are equal, bse will be equal.
Therefore, $\frac{n-1}{2}=5$
$\Rightarrow$ $n-1=5\times 2$
$\Rightarrow$ $n-1=10$
$\Rightarrow$ $n=10+1$
$\Rightarrow$ $n=11$