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 $r$. The terms are $a, ar, ar^2, \dots$.$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201}

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$150$

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(D) Given that $a_n$ is a $G$.$P$. with common ratio $r$. The terms are $a, ar, ar^2, \dots$.$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + \dots + a_{201} = ar^2 + ar^4 + \dots + ar^{200} = ar^2 \frac{(r^{200}-1)}{(r^2-1)} = 200$.$\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + \dots + a_{200} = ar + ar^3 + \dots + ar^{199} = ar \frac{(r^{200}-1)}{(r^2-1)} = 100$.Dividing the first equation by the second equation: $\frac{ar^2}{ar} = \frac{200}{100} \Rightarrow r = 2$.Now,$\sum_{n=1}^{200} a_n = a_1 + a_2 + \dots + a_{200} = a \frac{(r^{200}-1)}{(r-1)}$.From the second equation,$ar \frac{(r^{200}-1)}{(r^2-1)} = 100$. Since $r=2$,$2a \frac{(r^{200}-1)}{(2^2-1)} = 100 \Rightarrow 2a \frac{(r^{200}-1)}{3} = 100 \Rightarrow a \frac{(r^{200}-1)}{3} = 50 \Rightarrow a(r^{200}-1) = 150$.Substituting this into the sum formula: $\sum_{n=1}^{200} a_n = \frac{150}{2-1} = 150$.

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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