# What are next two terms of the GP $\sqrt{2},\;1/\sqrt{2},\;1/2\sqrt{2}$

$(a)\;\frac{1}{4\sqrt{4}},\frac{1}{6\sqrt{2}}\qquad(b)\;\frac{1}{\sqrt2},{\sqrt2}\qquad(c)\;\frac{1}{4\sqrt{2}},\frac{1}{8\sqrt{2}}\qquad(d)\;\frac{1}{4},\frac{1}{4\sqrt{2}}$

Answer : $\frac{1}{4\sqrt{2}},\frac{1}{8\sqrt{2}}$
Explanation : $r=\frac{T_{n}}{T_{n-1}}=\frac{1/\sqrt{2}}{\sqrt{2}}=\frac{1}{2}$
$next\;term=\frac{1}{2}*\frac{1}{2\sqrt{2}}=\frac{1}{4\sqrt{2}}$
$subsequent\;term=\frac{1}{2}*\frac{1}{4\sqrt{2}}=\frac{1}{8\sqrt{2}}.$