A G.P consists of 2n terms . If the sum of the terms . If the sum of the terms occupying the odd place is `S_(1),` and that of the terms in 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 of a geometric progression (G.P.) that consists of \(2n\) terms, given the sums of the terms in odd and even places, denoted as \(S_1\) and \(S_2\) respectively.

### Step-by-Step Solution:

1. **Identify the Terms of the G.P.**:
The terms of the G.P. can be represented as:
\[ a, ar, ar^2, ar^3, \ldots, ar^{2n-1} \]
where \(a\) is the first term and \(r\) is the common ratio.

2. **Sum of Terms in Odd Places**:
The terms in odd places are:
\[ a, ar^2, ar^4, \ldots, ar^{2n-2} \]
This forms a G.P. with the first term \(a\) and common ratio \(r^2\). The number of terms is \(n\).

The sum \(S_1\) of these terms can be calculated using the formula for the sum of a G.P.:
\[ S_1 = a \frac{1 - (r^2)^n}{1 - r^2} = a \frac{1 - r^{2n}}{1 - r^2} \]

3. **Sum of Terms in Even Places**:
The terms in even places are:
\[ ar, ar^3, ar^5, \ldots, ar^{2n-1} \]
This also forms a G.P. with the first term \(ar\) and common ratio \(r^2\). The number of terms is \(n\).

The sum \(S_2\) of these terms can be calculated similarly:
\[ S_2 = ar \frac{1 - (r^2)^n}{1 - r^2} = ar \frac{1 - r^{2n}}{1 - r^2} \]

4. **Divide \(S_2\) by \(S_1\)**:
Now, we can find the ratio \( \frac{S_2}{S_1} \):
\[ \frac{S_2}{S_1} = \frac{ar \frac{1 - r^{2n}}{1 - r^2}}{a \frac{1 - r^{2n}}{1 - r^2}} = \frac{ar}{a} = r \]

5. **Conclusion**:
Thus, we find that the common ratio \(r\) of the G.P. is given by:
\[ r = \frac{S_2}{S_1} \]

### Final Answer:
The common ratio of the progression is \( \frac{S_2}{S_1} \).