A geometric progression (GP) consists of 200 terms. If the sum of odd terms of the GP is m, and the sum of even terms of the GP is n. then what is its common ratio ?

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the terms of the GP**:
- Let the first term of the GP be \( a \) and the common ratio be \( r \).
- The terms of the GP can be represented as:
- 1st term: \( a \)
- 2nd term: \( ar \)
- 3rd term: \( ar^2 \)
- 4th term: \( ar^3 \)
- ...
- 200th term: \( ar^{199} \)

2. **Sum of the odd terms**:
- The odd terms are \( a, ar^2, ar^4, \ldots, ar^{198} \).
- This forms a new GP with the first term \( a \) and the common ratio \( r^2 \).
- The number of odd terms is 100 (from 1 to 199).
- The sum of the odd terms \( S_{odd} \) can be calculated using the formula for the sum of a GP:
\[ S_{odd} = a \frac{1 - (r^2)^{100}}{1 - r^2} = m \]

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

4. **Set up the ratio of the sums**:
- We have:
\[ \frac{S_{even}}{S_{odd}} = \frac{ar \frac{1 - (r^2)^{100}}{1 - r^2}}{a \frac{1 - (r^2)^{100}}{1 - r^2}} = \frac{ar}{a} = r \]
- Thus, we can express the ratio of the sums as:
\[ \frac{n}{m} = r \]

5. **Conclusion**:
- Therefore, the common ratio \( r \) is given by:
\[ r = \frac{n}{m} \]

### Final Answer:
The common ratio \( r \) of the geometric progression is \( \frac{n}{m} \).