A G.P consists of 200 terms. If the sum of odd terms of G.P is m and sum of even terms of G.P. is n, then the comon ratio of G.P is

Options

A. A. `m/n`
B. B. `n/m`
C. C. `m+n/m`
D. D. `n+m/n`

Answer

To solve the problem, we need to find the common ratio of a geometric progression (G.P.) that consists of 200 terms, given the sum of the odd terms is \( m \) and the sum of the even terms is \( n \).

### Step-by-Step Solution:

1. **Identify the terms of the G.P.**:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
\[ a, ar, ar^2, ar^3, ar^4, \ldots, ar^{199} \]

2. **Sum of the odd terms**:
The odd terms in the G.P. are:
\[ a, ar^2, ar^4, \ldots, ar^{198} \]
This forms another G.P. with the first term \( a \) and common ratio \( r^2 \). The number of odd terms is 100 (since there are 200 terms total).
The sum of the odd terms \( S_{\text{odd}} \) can be calculated using the formula for the sum of a finite G.P.:
\[ S_{\text{odd}} = \frac{a(1 - (r^2)^{100})}{1 - r^2} = m \]

3. **Sum of the even terms**:
The even terms in the G.P. are:
\[ ar, ar^3, ar^5, \ldots, ar^{199} \]
This also forms a G.P. with the first term \( ar \) and common ratio \( r^2 \). The number of even terms is also 100.
The sum of the even terms \( S_{\text{even}} \) can be calculated similarly:
\[ S_{\text{even}} = \frac{ar(1 - (r^2)^{100})}{1 - r^2} = n \]

4. **Set up the equations**:
We now have two equations:
\[ \frac{a(1 - (r^2)^{100})}{1 - r^2} = m \quad \text{(1)} \]
\[ \frac{ar(1 - (r^2)^{100})}{1 - r^2} = n \quad \text{(2)} \]

5. **Divide the two equations**:
By dividing equation (2) by equation (1), we eliminate \( a(1 - (r^2)^{100})/(1 - r^2) \):
\[ \frac{ar(1 - (r^2)^{100})}{a(1 - (r^2)^{100})} = \frac{n}{m} \]
This simplifies to:
\[ r = \frac{n}{m} \]

6. **Conclusion**:
The common ratio \( r \) of the G.P. is:
\[ r = \frac{n}{m} \]

### Final Answer:
The common ratio of the G.P. is \( \frac{n}{m} \).