A G.P. consist of 2n terms. If the sum of the terms occupying the odd places is S, and that of the terms occupying the even places is `S_(2)` then find the common ratio of the progression.

Answer

To solve the problem, we need to find the common ratio \( r \) of a geometric progression (G.P.) consisting of \( 2n \) terms, given the sums of the terms at odd and even positions.

### 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 at Odd Places**:
The terms occupying the odd places are:
\[ T_1, T_3, T_5, \ldots, T_{2n-1} \]
This can be expressed as:
\[ S_1 = A + Ar^2 + Ar^4 + \ldots + Ar^{2n-2} \]
This is a geometric series with the first term \( A \) and common ratio \( r^2 \). The number of terms is \( n \). The sum of this series can be calculated using the formula for the sum of a geometric series:
\[ S_1 = A \frac{1 - (r^2)^n}{1 - r^2} = A \frac{1 - r^{2n}}{1 - r^2} \]

3. **Sum of Terms at Even Places**:
The terms occupying the even places are:
\[ T_2, T_4, T_6, \ldots, T_{2n} \]
This can be expressed as:
\[ S_2 = Ar + Ar^3 + Ar^5 + \ldots + Ar^{2n-1} \]
This is also a geometric series with the first term \( Ar \) and common ratio \( r^2 \). The number of terms is \( n \). The sum of this series is:
\[ S_2 = Ar \frac{1 - (r^2)^n}{1 - r^2} = Ar \frac{1 - r^{2n}}{1 - r^2} \]

4. **Finding the Common Ratio**:
We have the two sums:
\[ S_1 = A \frac{1 - r^{2n}}{1 - r^2} \]
\[ S_2 = Ar \frac{1 - r^{2n}}{1 - r^2} \]
To find the common ratio \( r \), we can take the ratio of \( S_2 \) to \( S_1 \):
\[ \frac{S_2}{S_1} = \frac{Ar \frac{1 - r^{2n}}{1 - r^2}}{A \frac{1 - r^{2n}}{1 - r^2}} = r \]
Hence, we find that:
\[ r = \frac{S_2}{S_1} \]

### Final Answer:
The common ratio \( r \) of the progression is:
\[ r = \frac{S_2}{S_1} \]