Let `a_n` is a positive term of a GP and `sum_(n=1)^100 a_(2n + 1)= 200, sum_(n=1)^100 a_(2n) = 200`, find `sum_(n=1)^200 a_(2n) =` ?

Options

A. 300
B. 150
C. 175
D. 225

Answer

To solve the problem, we need to find the value of \( \sum_{n=1}^{200} a_{2n} \) given that \( \sum_{n=1}^{100} a_{2n+1} = 200 \) and \( \sum_{n=1}^{100} a_{2n} = 200 \).

### Step-by-Step Solution:

1. **Understanding the terms of the GP**:
Let \( a_n \) be the \( n \)-th term of a geometric progression (GP). We can express the terms as:
\[ a_n = ar^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Sum of odd indexed terms**:
The odd indexed terms are \( a_1, a_3, a_5, \ldots, a_{201} \) which can be expressed as:
\[ \sum_{n=1}^{100} a_{2n-1} = a + ar^2 + ar^4 + \ldots + ar^{200} \]
This is a geometric series with first term \( a \) and common ratio \( r^2 \):
\[ \sum_{n=1}^{100} a_{2n-1} = a \frac{(r^2)^{100} - 1}{r^2 - 1} = 200 \]

3. **Sum of even indexed terms**:
The even indexed terms are \( a_2, a_4, a_6, \ldots, a_{200} \):
\[ \sum_{n=1}^{100} a_{2n} = ar + ar^3 + ar^5 + \ldots + ar^{199} \]
This is also a geometric series with first term \( ar \) and common ratio \( r^2 \):
\[ \sum_{n=1}^{100} a_{2n} = ar \frac{(r^2)^{100} - 1}{r^2 - 1} = 200 \]

4. **Setting up the equations**:
We have two equations:
\[ a \frac{r^{200} - 1}{r^2 - 1} = 200 \quad \text{(1)} \]
\[ ar \frac{r^{200} - 1}{r^2 - 1} = 200 \quad \text{(2)} \]

5. **Dividing the equations**:
Dividing equation (2) by equation (1):
\[ \frac{ar \frac{r^{200} - 1}{r^2 - 1}}{a \frac{r^{200} - 1}{r^2 - 1}} = \frac{200}{200} \]
This simplifies to:
\[ r = 1 \]

6. **Finding the value of \( a \)**:
Substitute \( r = 2 \) into either equation. Using equation (1):
\[ a \frac{2^{200} - 1}{2^2 - 1} = 200 \]
Simplifying gives:
\[ a \frac{2^{200} - 1}{3} = 200 \]
Thus,
\[ a(2^{200} - 1) = 600 \quad \text{(3)} \]

7. **Finding \( \sum_{n=1}^{200} a_{2n} \)**:
Now we want to find:
\[ \sum_{n=1}^{200} a_{2n} = ar \frac{(r^2)^{100} - 1}{r^2 - 1} \]
Substituting \( r = 2 \):
\[ = 2a \frac{2^{200} - 1}{3} \]
From equation (3), we can find \( a \):
\[ \sum_{n=1}^{200} a_{2n} = 2 \cdot \frac{600}{2^{200} - 1} \cdot \frac{2^{200} - 1}{3} = 400 \]

### Final Answer:
\[ \sum_{n=1}^{200} a_{2n} = 400 \]