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 $$r$$.
Since all terms are positive, we know $$a > 0$$ and $$r > 0$$.

1. Set up the first equation

The first sum is for the odd positioned terms starting from $$n=1$$:

$$\sum_{n=1}^{100} a_{2n+1} = a_3 + a_5 + a_7 + \dots + a_{201} = 200$$

Expressing this in terms of $$a$$ and $$r$$:

$$ar^2 + ar^4 + ar^6 + \dots + ar^{200} = 200$$

This series is a G.P. itself with $$100$$ terms, a first term of $$ar^2$$, and a common ratio of $$r^2$$.
Apply the standard sum formula for a G.P.:

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

$$\frac{ar^2(r^{200} - 1)}{r^2 - 1} = 200 \quad \text{(Equation 1)}$$

2. Set up the second equation

The second sum is for the even positioned terms starting from $$n=1$$:

$$\sum_{n=1}^{100} a_{2n} = a_2 + a_4 + a_6 + \dots + a_{200} = 100$$

Expressing this in terms of $$a$$ and $$r$$:

$$ar + ar^3 + ar^5 + \dots + ar^{199} = 100$$

This series is also a G.P. with $$100$$ terms, a first term of $$ar$$, and a common ratio of $$r^2$$.
Apply the sum formula:

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

$$\frac{ar(r^{200} - 1)}{r^2 - 1} = 100 \quad \text{(Equation 2)}$$

3. Solve for r and a

Divide Equation 1 by Equation 2 to eliminate $$a$$ and the complex brackets:

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

All the common terms cancel out perfectly, leaving just:

$$r = 2$$

Now substitute $$r = 2$$ back into Equation 2 to find $$a$$:

$$\frac{a(2)(2^{200} - 1)}{2^2 - 1} = 100$$

$$\frac{2a(2^{200} - 1)}{3} = 100$$

Multiply both sides by 3 and divide by 2:

$$a(2^{200} - 1) = 150$$

$$a = \frac{150}{2^{200} - 1}$$

4. Calculate the final requested sum

We need to find the sum of the first 200 terms of the original G.P.:

$$S_{200} = \sum_{n=1}^{200} a_n = a + ar + ar^2 + \dots + ar^{199}$$

Using the standard sum formula for $$200$$ terms:

$$S_{200} = \frac{a(r^{200} - 1)}{r - 1}$$

Substitute the values of $$a$$ and $$r$$ we just found:

$$S_{200} = \frac{\left(\frac{150}{2^{200} - 1}\right)(2^{200} - 1)}{2 - 1}$$

The identical brackets in the numerator and denominator cancel out, and the denominator simplifies to 1:

$$S_{200} = \frac{150}{1} = 150$$