A G.P. consists of 2n terms. If the sum of the terms occupying the odd places is `S_(1)`, and that of the terms in the even places is `S_(2)`, then `S_(2)/S_(1)`, is

Options

A. independent of a
B. independent of r
C. independent of a and r
D. dependent on r

Explanation

To solve the problem, we need to find the ratio \( \frac{S_2}{S_1} \) where \( S_1 \) is the sum of the terms in the odd places and \( S_2 \) is the sum of the terms in the even places of a geometric progression (G.P.) consisting of \( 2n \) terms.

### Step-by-Step Solution:

1. **Identify the Terms of the G.P.**:
The terms of the G.P. can be expressed as:
\[ T_1 = A, \quad T_2 = AR, \quad T_3 = AR^2, \quad \ldots, \quad T_{2n} = AR^{2n-1} \]

2. **Sum of Terms in Odd Places (\( S_1 \))**:
The terms in odd places are:
\[ T_1, T_3, T_5, \ldots, T_{2n-1} \]
This corresponds to:
\[ S_1 = A + AR^2 + AR^4 + \ldots + AR^{2n-2} \]
This is a geometric series with the first term \( A \) and common ratio \( R^2 \). The number of terms is \( n \).
The sum of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Applying this formula:
\[ S_1 = A \frac{1 - (R^2)^n}{1 - R^2} = A \frac{1 - R^{2n}}{1 - R^2} \]

3. **Sum of Terms in Even Places (\( S_2 \))**:
The terms in even places are:
\[ T_2, T_4, T_6, \ldots, T_{2n} \]
This corresponds to:
\[ S_2 = AR + AR^3 + AR^5 + \ldots + AR^{2n-1} \]
This is also a geometric series with the first term \( AR \) and common ratio \( R^2 \). The number of terms is \( n \).
Thus:
\[ S_2 = AR \frac{1 - (R^2)^n}{1 - R^2} = AR \frac{1 - R^{2n}}{1 - R^2} \]

4. **Finding the Ratio \( \frac{S_2}{S_1} \)**:
Now, we can find the ratio:
\[ \frac{S_2}{S_1} = \frac{AR \frac{1 - R^{2n}}{1 - R^2}}{A \frac{1 - R^{2n}}{1 - R^2}} \]
The terms \( A \) and \( \frac{1 - R^{2n}}{1 - R^2} \) cancel out:
\[ \frac{S_2}{S_1} = R \]

### Conclusion:
Thus, the final result is:
\[ \frac{S_2}{S_1} = R \]