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Let $a_{n}$ be the $n^{\text {th }}$ term of a G.P. of positive terms.
If $\sum\limits_{n=1}^{100} a_{2 n+1}=200$ and $\sum\limits_{n=1}^{100} a_{2 n}=100,$ then $\sum\limits_{n=1}^{200} a_{n}$ is equal to
$225$
$175$
$300$
$150$
Solution
$\sum_{n=1}^{100} a_{2 n+1}=200 \Rightarrow a_{3}+a_{5}+a_{7}+\ldots .+a_{201}=200$
$\Rightarrow \operatorname{ar}^{2} \frac{\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=200$
$\sum_{n=1}^{100} a_{2 n}=100 \Rightarrow a_{2}+a_{4}+a_{6}+\ldots+a_{200}=100$
$\Rightarrow \frac{\operatorname{ar}\left(\mathrm{r}^{200}-1\right)}{\left(\mathrm{r}^{2}-1\right)}=100$
On dividing $\mathrm{r}=2$
on adding $a_{2}+a_{3}+a_{4}+a_{5}+\ldots+a_{200}+a_{201}=300$
$\Rightarrow \mathrm{r}\left(\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\ldots .+\mathrm{a}_{200}\right)=300$
$\Rightarrow \sum_{n=1}^{200} a_{n}=150$
