The given G.P. is \(\sqrt{7},\sqrt{21},3\sqrt{7},\) ...
Here, a = \(\sqrt{7}\)
r = \(\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\)
Sn = \(\frac{a(1-r^n)}{1-r}\)
∴ Sn = \(\frac{\sqrt7[1-({\sqrt3})n]}{1-\sqrt3}\)
= \(\frac{\sqrt{7}[1-(\sqrt{3})n]}{1-\sqrt{3}}\times\frac{1+\sqrt{3}}{1+\sqrt{3}}\) (By rationalizing)
= \(\frac{\sqrt{7}(1+\sqrt{3})[1-(\sqrt{3}n]}{1-3}\)
= \(-\frac{\sqrt{7}(1+\sqrt{3})}{2[\frac{1-(3)n}{2}]}\)
= \(\frac{\sqrt{7}(1+\sqrt{3})}{2\bigg[\frac{(3)n}{2}{-1}\bigg]}\)
(a) 2 ,\(2\sqrt{2}\) , 4 ,.... is 128 ?
(b) \(\sqrt{3}\), 3, \(3\sqrt{3}\),... is 729 ?
(c) \(\frac{1}{3},\frac{1}{9},\frac{1}{27}\) ,.... is \(\frac{1}{19683}\) ?