Let the G.P. have first term $a$ and common ratio $r$.

Then $a_n = ar^{n-1}$ with $a>0$ and, since all terms are positive, $r>0$.

Sum of even-indexed terms from $a_2$ to $a_{200}$:

$\sum_{n=1}^{100} a_{2n} = a_2+a_4+\cdots+a_{200}$

Now, $a_{2n}=ar^{2n-1}=ar(r^2)^{n-1}$

So this is a G.P. with first term $ar$ and ratio $r^2$. Hence, $\sum_{n=1}^{100} a_{2n} = ar\left(\frac{1-r^{200}}{1-r^2}\right)=100 \quad ...(1)$

Sum of odd-indexed terms from $a_3$ to $a_{201}$:

$\sum_{n=1}^{100} a_{2n+1}=a_3+a_5+\cdots+a_{201}$

Here, $a_{2n+1}=ar^{2n}=ar^2(r^2)^{n-1}$

So, $\sum_{n=1}^{100} a_{2n+1}=ar^2\left(\frac{1-r^{200}}{1-r^2}\right)=200 \quad ...(2)$

Divide equation (2) by equation (1):

$\frac{ar^2\left(\frac{1-r^{200}}{1-r^2}\right)}{ar\left(\frac{1-r^{200}}{1-r^2}\right)}=\frac{200}{100}$

$r=2$

But if $r=2$, then the G.P. terms are increasing, and the sum of odd terms from $a_3$ onward being twice the even terms from $a_2$ onward is consistent. However, let us directly relate the two sums termwise:

Since $a_{2n+1}=r\,a_{2n}$ and the two sums are over corresponding 100 terms, $\sum_{n=1}^{100} a_{2n+1}=r\sum_{n=1}^{100} a_{2n}$

Hence, $200=r\cdot 100 \implies r=2$

Now we need $\sum_{n=1}^{200} a_n = \sum_{n=1}^{100} a_{2n-1} + \sum_{n=1}^{100} a_{2n}$

We are given only $\sum_{n=1}^{100} a_{2n}=100$ and $\sum_{n=1}^{100} a_{2n+1}=200$.

Observe that $\sum_{n=1}^{100} a_{2n+1} = a_3+a_5+\cdots+a_{201}$ and $\sum_{n=1}^{100} a_{2n-1} = a_1+a_3+\cdots+a_{199}$

These differ by removing $a_{201}$ and adding $a_1$: $\sum_{n=1}^{100} a_{2n-1} = \sum_{n=1}^{100} a_{2n+1} + a_1 - a_{201}$

Since $r=2$, $a_{201}$ is huge compared to $a_1$, so this cannot lead to one of the small options unless the intended question was likely $\sum_{n=1}^{100} a_{2n-1}=200, \quad \sum_{n=1}^{100} a_{2n}=100$ in which case $\sum_{n=1}^{200} a_n=200+100=300.$

Therefore, from the data as written, the consistent direct conclusion is that the odd-indexed shifted sum is $200$, even-indexed sum is $100$, giving ratio $r=2$, but then none of the options fit naturally for $\sum_{n=1}^{200} a_n$ unless there is a typo.

If one interprets the intended first sum as the sum of odd terms among the first 200 terms, then $\sum_{n=1}^{200} a_n = \sum_{n=1}^{100} a_{2n-1}+\sum_{n=1}^{100} a_{2n}=200+100=300.$