Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms.
If $\sum_{n=1}^{100} a_{2 n+1}=200$ and
$\sum_{n=1}^{100} a_{2 n}=100$, then $\sum_{n=1}^{200} a_{n}$ is equal to :
Correct Option: , 4
Let G.P. be $a, a r, a r^{2}$
$\sum_{n=1}^{100} a_{2 n+1}=a_{3}+a_{5}+\ldots . .+a_{201}=200$
$\Rightarrow \frac{a r^{2}\left(r^{200}-1\right)}{r^{2}-1}=200$ ...(i)
$\sum_{n=1}^{100} a_{2 n}=a_{2}+a_{4}+\ldots . .+a_{200}=100$
$\Rightarrow \quad \frac{\operatorname{ar}\left(r^{200}-1\right)}{r^{2}-1}=100$...(ii)
From equations (i) and (ii), $r=2$ and
$a_{2}+a_{3}+\ldots . .+a_{200}+a_{201}=300$
$\Rightarrow r\left(a_{1}+\ldots . .+a_{200}\right)=300$
$\Rightarrow \quad \sum_{n=1}^{200} a_{n}=\frac{300}{r}=150$
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