Let a_{n} be the n^{text{th}} term of a G.P. | vedclass

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(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$. $\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = 200$ This is a $G$.$P$. with firs

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$150$

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(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$. $\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = 200$ This is a $G$.$P$. with first term $a_3 = ar^2$ and common ratio $r^2$ with $100$ terms. So,$ar^2 \frac{(r^2)^{100} - 1}{r^2 - 1} = ar^2 \frac{r^{200} - 1}{r^2 - 1} = 200$ $\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200} = 100$ This is a $G$.$P$. with first term $a_2 = ar$ and common ratio $r^2$ with $100$ terms. So,$ar \frac{(r^2)^{100} - 1}{r^2 - 1} = ar \frac{r^{200} - 1}{r^2 - 1} = 100$ Dividing the two equations: $\frac{ar^2 \frac{r^{200} - 1}{r^2 - 1}}{ar \frac{r^{200} - 1}{r^2 - 1}} = \frac{200}{100} \Rightarrow r = 2$ We need to find $S = \sum_{n=1}^{200} a_n = a_1 + a_2 + \dots + a_{200}$. Note that $\sum_{n=1}^{100} a_{2n+1} + \sum_{n=1}^{100} a_{2n} = a_2 + a_3 + a_4 + \dots + a_{201} = 300$. Since $a_{k+1} = r a_k$,we have $a_2 + a_3 + \dots + a_{201} = r(a_1 + a_2 + \dots + a_{200}) = 300$. Substituting $r = 2$: $2 \sum_{n=1}^{200} a_n = 300 \Rightarrow \sum_{n=1}^{200} a_n = 150$.

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

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