The number of terms of a G.P. are even. If the sum of all terms of the series is 5 times the sum of all terms at odd positions, then find the common ratio.

Explanation

To solve the problem, we need to find the common ratio \( r \) of a geometric progression (G.P.) given that the sum of all terms is 5 times the sum of the terms at odd positions.

### Step-by-Step Solution:

1. **Define the G.P. Terms**:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be represented as:
\[ a, ar, ar^2, ar^3, \ldots, ar^{n-1} \]
where \( n \) is the total number of terms, and since \( n \) is even, we can write \( n = 2m \) for some integer \( m \).

2. **Sum of All Terms**:
The sum \( S \) of the first \( n \) terms of a G.P. is given by:
\[ S = \frac{a(r^n - 1)}{r - 1} \]
Substituting \( n = 2m \):
\[ S = \frac{a(r^{2m} - 1)}{r - 1} \]

3. **Sum of Terms at Odd Positions**:
The terms at odd positions are:
\[ a, ar^2, ar^4, \ldots, ar^{2(m-1)} \]
This is also a G.P. with first term \( a \) and common ratio \( r^2 \). The number of odd-positioned terms is \( m \). Thus, the sum of the odd-positioned terms \( S_{\text{odd}} \) is:
\[ S_{\text{odd}} = \frac{a(r^{2m} - 1)}{r^2 - 1} \]

4. **Set Up the Equation**:
According to the problem, the sum of all terms is 5 times the sum of the terms at odd positions:
\[ S = 5 S_{\text{odd}} \]
Substituting the expressions for \( S \) and \( S_{\text{odd}} \):
\[ \frac{a(r^{2m} - 1)}{r - 1} = 5 \cdot \frac{a(r^{2m} - 1)}{r^2 - 1} \]

5. **Cancel \( a(r^{2m} - 1) \)**:
Assuming \( a(r^{2m} - 1) \neq 0 \), we can cancel it from both sides:
\[ \frac{1}{r - 1} = 5 \cdot \frac{1}{r^2 - 1} \]

6. **Cross-Multiply**:
Cross-multiplying gives:
\[ r^2 - 1 = 5(r - 1) \]
Expanding the right side:
\[ r^2 - 1 = 5r - 5 \]

7. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ r^2 - 5r + 4 = 0 \]

8. **Factoring the Quadratic**:
Factoring the quadratic equation:
\[ (r - 4)(r - 1) = 0 \]

9. **Finding the Values of \( r \)**:
Thus, the solutions for \( r \) are:
\[ r = 4 \quad \text{or} \quad r = 1 \]

10. **Conclusion**:
Since the problem states that the number of terms is even and the sum of the terms at odd positions must be less than the total sum, we discard \( r = 1 \) (as it would make all terms equal). Therefore, the common ratio is:
\[ r = 4 \]