Alg8.12

Which term of the Geometri Progression(G.P) $2,2\sqrt{2},4......\,$ is $64\,$

Here $a=2,r=\frac{2\sqrt{2}}{2}=\sqrt{2}\,$

Now $t_n=2(\sqrt{2})^{n-1}=64\,$

Simplifying $(\sqrt{2})^{n-1}=32\,$

$2^{\frac{n-1}{2}}=2^5\,$

Equating the indices we get

$\frac{n-1}{2}=5,n=11\,$

Thus $64\,$ is the 11th term of GP.