The sum of first 12 terms of a G.P. is five times the sum of the first 6 terms. Find the common ratio.

Answer

To solve the problem, we need to find the common ratio \( r \) of a geometric progression (G.P.) given that the sum of the first 12 terms is five times the sum of the first 6 terms.

### Step-by-Step Solution:

1. **Formula for the Sum of Terms in a G.P.**:
The sum of the first \( n \) terms of a G.P. is given by the formula:
\[ S_n = \frac{A(r^n - 1)}{r - 1} \]
where \( A \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

2. **Expressing the Sums**:
For the first 12 terms:
\[ S_{12} = \frac{A(r^{12} - 1)}{r - 1} \]
For the first 6 terms:
\[ S_{6} = \frac{A(r^{6} - 1)}{r - 1} \]

3. **Setting Up the Equation**:
According to the problem, the sum of the first 12 terms is five times the sum of the first 6 terms:
\[ S_{12} = 5 S_{6} \]
Substituting the expressions for \( S_{12} \) and \( S_{6} \):
\[ \frac{A(r^{12} - 1)}{r - 1} = 5 \cdot \frac{A(r^{6} - 1)}{r - 1} \]

4. **Cancelling Common Terms**:
Since \( A \) and \( r - 1 \) are common in both sides (assuming \( r \neq 1 \)), we can cancel them:
\[ r^{12} - 1 = 5(r^{6} - 1) \]

5. **Expanding the Equation**:
Expanding the right side:
\[ r^{12} - 1 = 5r^{6} - 5 \]
Rearranging gives:
\[ r^{12} - 5r^{6} + 4 = 0 \]

6. **Substituting \( x = r^{6} \)**:
Let \( x = r^{6} \). The equation becomes:
\[ x^2 - 5x + 4 = 0 \]

7. **Factoring the Quadratic**:
Factoring the quadratic equation:
\[ (x - 4)(x - 1) = 0 \]
Thus, \( x = 4 \) or \( x = 1 \).

8. **Finding \( r \)**:
- If \( x = 4 \):
\[ r^{6} = 4 \implies r = 4^{1/6} = 2^{2/6} = 2^{1/3} \]
- If \( x = 1 \):
\[ r^{6} = 1 \implies r = 1 \]
However, since \( r \) cannot be 1 in a G.P., we discard this solution.

9. **Final Result**:
The common ratio \( r \) is:
\[ r = 2^{1/3} \]