Let $$a_n$$ be the $$n^{th}$$ 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:

Let the first term of the G.P. be $$a$$ and the common ratio be $$r$$. Hence every term can be written as $$a_n=a\,r^{\,n-1}$$.

We are told that

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

First we translate these two summations into algebraic form. For the even-indexed terms we have

$$a_{2n}=a\,r^{(2n)-1}=a\,r^{2n-1} =a\,r\,(r^2)^{\,n-1}.$$ So

$$\sum_{n=1}^{100} a_{2n} =a\,r\sum_{n=1}^{100}(r^2)^{\,n-1} =a\,r\;\frac{1-(r^2)^{100}}{1-r^2} =\frac{a\,r\left(1-r^{200}\right)}{1-r^2} =100.$$

For the odd-indexed terms (beginning with the third term) we have

$$a_{2n+1}=a\,r^{(2n+1)-1}=a\,r^{2n} =a\,r^{2}(r^2)^{\,n-1},$$ so

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

Dividing the second equation by the first eliminates the common factor $$a\,\dfrac{1-r^{200}}{1-r^2}$$ and gives

$$\frac{200}{100}=\frac{a\,r^{2}\left(1-r^{200}\right)/(1-r^2)} {a\,r \left(1-r^{200}\right)/(1-r^2)} =\frac{r^{2}}{r}=r.$$

Thus $$r=2.$$ (All terms are positive, so the positive value is taken.)

Now substitute $$r=2$$ in the equation for the even-indexed sum:

$$\frac{a\,(2)\,\bigl(1-2^{200}\bigr)}{1-4}=100 \;\Longrightarrow\; \frac{2a\,(1-2^{200})}{-3}=100 \;\Longrightarrow\; \frac{2a\,(2^{200}-1)}{3}=100 \;\Longrightarrow\; a=\frac{150}{2^{200}-1}.$$

Our goal is $$S=\sum_{n=1}^{200} a_n.$$ Using the standard sum formula for a G.P.,

$$S=\frac{a\,(1-r^{200})}{1-r} =\frac{a\,(1-2^{200})}{1-2} =\frac{a\,(1-2^{200})}{-1} =a\,(2^{200}-1).$$

Insert the value of $$a$$ obtained above:

$$S=\frac{150}{2^{200}-1}\;(2^{200}-1)=150.$$

Therefore the sum of the first 200 terms is $$150.$$

Hence, the correct answer is Option D.