Question:
Find the sum of 10 terms of the geometric series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ldots$
Solution:
We need to find the sum of 10 terms of GP.
Sum of n terms of GP, with first term, a, common ratio, r,
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$
So, the sum of given GP up to 10 terms, with $\mathrm{a}=\sqrt{2}$,
$r=\sqrt{3}, n=10$
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{\sqrt{2}\left((\sqrt{3})^{10}-1\right)}{\sqrt{3}-1}$
The requires sum, $S_{n}=\frac{\sqrt{2\left((3)^{5}-1\right)}}{\sqrt{3}-1}=467.5$