Let `a_n` be the nth therm of a G.P of positive numbers .Let `Sigma_(n=1)^(100) a_(2n)=alpha and Sigma_(n=1)^(100)a_(an-1)=beta` then the common ratio is

Options

A. `alpha//beta`
B. `beta//alpha`
C. `sqrt(alpha//beta)`
D. `sqrt(beta//alpha)`

Explanation

To solve the problem, we need to find the common ratio of a geometric progression (G.P.) given the sums of certain terms. Let's denote the first term of the G.P. as \( a \) and the common ratio as \( r \).

### Step-by-Step Solution:

1. **Identify the terms in the G.P.**:
The \( n \)-th term of the G.P. can be expressed as:
\[ a_n = a r^{n-1} \]

2. **Calculate \( \alpha = \sum_{n=1}^{100} a_{2n} \)**:
The terms \( a_{2n} \) correspond to the terms of the G.P. at even indices:
\[ a_{2n} = a r^{2n-1} \]
Therefore, we can express \( \alpha \) as:
\[ \alpha = \sum_{n=1}^{100} a r^{2n-1} = a \sum_{n=1}^{100} r^{2n-1} \]
This is a geometric series with first term \( r \) and common ratio \( r^2 \):
\[ \alpha = a r \sum_{n=0}^{99} (r^2)^n = a r \cdot \frac{1 - (r^2)^{100}}{1 - r^2} = a r \cdot \frac{1 - r^{200}}{1 - r^2} \]

3. **Calculate \( \beta = \sum_{n=1}^{100} a_{2n-1} \)**:
The terms \( a_{2n-1} \) correspond to the terms of the G.P. at odd indices:
\[ a_{2n-1} = a r^{2n-2} \]
Therefore, we can express \( \beta \) as:
\[ \beta = \sum_{n=1}^{100} a r^{2n-2} = a \sum_{n=1}^{100} r^{2n-2} = a \sum_{n=0}^{99} r^{2n} = a \cdot \frac{1 - r^{200}}{1 - r^2} \]

4. **Relate \( \alpha \) and \( \beta \)**:
Now we have:
\[ \alpha = a r \cdot \frac{1 - r^{200}}{1 - r^2} \]
\[ \beta = a \cdot \frac{1 - r^{200}}{1 - r^2} \]
Dividing \( \alpha \) by \( \beta \):
\[ \frac{\alpha}{\beta} = \frac{a r \cdot \frac{1 - r^{200}}{1 - r^2}}{a \cdot \frac{1 - r^{200}}{1 - r^2}} = r \]

5. **Conclusion**:
Thus, the common ratio \( r \) can be expressed as:
\[ r = \frac{\alpha}{\beta} \]

### Final Answer:
The common ratio \( r \) is given by:
\[ r = \frac{\alpha}{\beta} \]