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Q. Let $a_n$ be the $n^{th}$ term of a G.P. of positive terms. If $x=\displaystyle \sum_{n=1}^{100}a_{2n+1}=200$ and $x=\displaystyle \sum_{n=1}^{100}a_{2n}=100$, $x=\displaystyle \sum_{n=1}^{200}a_n$ is equal to :

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Solution:

$\sum\limits^{100}_{n = 1} a_{2n+1} = 200 \Rightarrow a_{3} + a_{5} + a_{7} + .... + a_{201} = 200$
$\Rightarrow ar^{2} \frac{\left(r^{200}-1\right)}{\left(r^{2}-1\right)} = 200$
$\sum\limits^{100}_{n = 1} a_{2n} = 100 \Rightarrow a_{2} + a_{4} + a_{6} + ... + a_{200} = 100$
$\Rightarrow \frac{ar\left(r^{200}-1\right)}{\left(r^{2}-1\right)} = 100$
On dividing $r = 2$
on adding $a_{2} + a_{3} + a_{4} + a_{5} + ... +a_{200} + a_{201} = 300$
$\Rightarrow r\left(a_{1} + a_{2} + a_{3} + .... + a_{200}\right) = 300$
$\Rightarrow \sum\limits^{100}_{n = 1} a_{n} = 150$