Which term of the sequence $2,2 \sqrt{2}, 4, \ldots$ is 128 ?

Question

Professor PC Thomas Classes · Answer

$a=2$, $r=\sqrt{2}$. $2(\sqrt{2})^{n-1} = 128 \implies (\sqrt{2})^{n-1} = 64 = 2^6 = (\sqrt{2})^{12} \implies n-1=12 \implies n=13$.