1. Key Concepts and Formulas

• Geometric Progression (G.P.): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The $n^{\text{th}}$ term is $a_n = ar^{n-1}$.
• Sum of a G.P.: The sum of the first $N$ terms is $S_N = \frac{a(r^N - 1)}{r - 1}$ (for $r \neq 1$).
• Properties of Sums: Sums of specific terms within a G.P. can themselves form a G.P.

2. Step-by-Step Solution

Step 1: Understand the Given Information and Define the G.P. We are given a G.P. of positive terms, denoted by $a_n$. Let the first term be $a$ and the common ratio be $r$. Since all terms are positive, $a > 0$ and $r > 0$. The given information is: $\sum_{n=1}^{100} a_{2n+1} = 200 \quad \text{...(1)}$ $\sum_{n=1}^{100} a_{2n} = 100 \quad \text{...(2)}$ We need to find $\sum_{n=1}^{200} a_n$.

Step 2: Analyze the Sum of Odd-Indexed Terms The sum $\sum_{n=1}^{100} a_{2n+1}$ represents the sum of terms $a_3, a_5, a_7, \dots, a_{201}$. These terms can be written as: $a_3 = ar^{2}$ $a_5 = ar^{4}$ $a_7 = ar^{6}$ ... $a_{201} = ar^{200}$ This sequence $ar^2, ar^4, ar^6, \dots, ar^{200}$ is itself a G.P. with:

• First term ($A_1$) = $ar^2$
• Common ratio ($R$) = $\frac{ar^4}{ar^2} = r^2$
• Number of terms ($N_1$) = 100 (since $n$ goes from 1 to 100) Using the sum formula for this new G.P.: $\sum_{n=1}^{100} a_{2n+1} = \frac{A_1(R^{N_1} - 1)}{R - 1} = \frac{ar^2((r^2)^{100} - 1)}{r^2 - 1} = \frac{ar^2(r^{200} - 1)}{r^2 - 1}$ From the given information (1), we have: $\frac{ar^2(r^{200} - 1)}{r^2 - 1} = 200 \quad \text{...(1')}$

Step 3: Analyze the Sum of Even-Indexed Terms The sum $\sum_{n=1}^{100} a_{2n}$ represents the sum of terms $a_2, a_4, a_6, \dots, a_{200}$. These terms can be written as: $a_2 = ar^{1}$ $a_4 = ar^{3}$ $a_6 = ar^{5}$ ... $a_{200} = ar^{199}$ This sequence $ar, ar^3, ar^5, \dots, ar^{199}$ is also a G.P. with:

• First term ($A_2$) = $ar$
• Common ratio ($R$) = $\frac{ar^3}{ar} = r^2$
• Number of terms ($N_2$) = 100 (since $n$ goes from 1 to 100) Using the sum formula for this new G.P.: $\sum_{n=1}^{100} a_{2n} = \frac{A_2(R^{N_2} - 1)}{R - 1} = \frac{ar((r^2)^{100} - 1)}{r^2 - 1} = \frac{ar(r^{200} - 1)}{r^2 - 1}$ From the given information (2), we have: $\frac{ar(r^{200} - 1)}{r^2 - 1} = 100 \quad \text{...(2')}$

Step 4: Determine the Common Ratio ($r$) of the Original G.P. We have two equations (1') and (2') involving $a$ and $r$. To find $r$, we can divide equation (1') by equation (2'). Since $a>0$ and $r>0$, and the sums are non-zero, we know $r \neq 1$ (otherwise $r^2-1=0$ and $a_n$ would be constant, leading to a contradiction $100a=200$ and $100a=100$). Also, $r^{200}-1 \neq 0$. Thus, we can safely cancel common terms. $\frac{\frac{ar^2(r^{200} - 1)}{r^2 - 1}}{\frac{ar(r^{200} - 1)}{r^2 - 1}} = \frac{200}{100}$ $\frac{ar^2}{ar} = 2$ $r = 2$ The common ratio of the original G.P. is $r=2$.

Step 5: Calculate the Sum of the First 200 Terms We need to find $\sum_{n=1}^{200} a_n$, which is the sum of the first 200 terms of the original G.P. $S_{200} = \frac{a(r^{200} - 1)}{r - 1}$ Substitute $r=2$: $S_{200} = \frac{a(2^{200} - 1)}{2 - 1} = a(2^{200} - 1)$ To find the value of $a(2^{200} - 1)$, we can use equation (2') and substitute $r=2$: $\frac{ar(r^{200} - 1)}{r^2 - 1} = 100$ $\frac{a(2)(2^{200} - 1)}{2^2 - 1} = 100$ $\frac{2a(2^{200} - 1)}{4 - 1} = 100$ $\frac{2a(2^{200} - 1)}{3} = 100$ Multiply both sides by 3: $2a(2^{200} - 1) = 300$ Divide by 2: $a(2^{200} - 1) = 150$ Since $S_{200} = a(2^{200} - 1)$, we have: $S_{200} = 150$

3. Common Mistakes & Tips

• Incorrectly identifying sub-G.P. parameters: Ensure the first term and common ratio of the sub-G.P.s (odd and even indexed terms) are correctly derived. The common ratio of these sub-G.P.s is $r^2$, not $r$.
• Algebraic errors when dividing sums: Be careful when cancelling terms during the division of equations. Verify that all cancelled terms are non-zero.
• Directly relating sums: Avoid trying to directly relate the given sums to $S_{200}$ without first finding the common ratio $r$. The strategy of finding $r$ by dividing the two given sums is crucial.

4. Summary The problem involves a Geometric Progression where we are given sums of alternating sets of terms. By recognizing that the sums of odd-indexed terms and even-indexed terms also form Geometric Progressions with a common ratio of $r^2$, we can set up two equations. Dividing these equations allows us to efficiently find the common ratio $r$ of the original G.P. Once $r$ is known, we substitute it back into one of the sum equations to determine the value of the expression $a(r^{200}-1)$, which is precisely the sum of the first 200 terms.

5. Final Answer The final answer is $\boxed{150}$.