If the sum of first 10 terms is 33 times the sum of first 5 terms of G.P. find the common ratio.

Answer

To solve the problem, we need to find the common ratio of a geometric progression (G.P.) given that the sum of the first 10 terms is 33 times the sum of the first 5 terms.

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
Let the first term of the G.P. be \( a \) and the common ratio be \( r \).

2. **Write the formulas for the sums**:
The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \quad (r \neq 1) \]
Therefore, the sum of the first 10 terms \( S_{10} \) and the sum of the first 5 terms \( S_5 \) can be expressed as:
\[ S_{10} = a \frac{1 - r^{10}}{1 - r} \]
\[ S_5 = a \frac{1 - r^5}{1 - r} \]

3. **Set up the equation based on the problem statement**:
According to the problem, the sum of the first 10 terms is 33 times the sum of the first 5 terms:
\[ S_{10} = 33 S_5 \]
Substituting the expressions for \( S_{10} \) and \( S_5 \):
\[ a \frac{1 - r^{10}}{1 - r} = 33 \left( a \frac{1 - r^5}{1 - r} \right) \]

4. **Simplify the equation**:
Since \( a \) and \( (1 - r) \) are common on both sides (assuming \( a \neq 0 \) and \( r \neq 1 \)), we can cancel them out:
\[ 1 - r^{10} = 33(1 - r^5) \]

5. **Rearranging the equation**:
Expanding the right side:
\[ 1 - r^{10} = 33 - 33r^5 \]
Rearranging gives:
\[ r^{10} - 33r^5 + 32 = 0 \]

6. **Let \( x = r^5 \)**:
This substitution transforms the equation into a quadratic:
\[ x^2 - 33x + 32 = 0 \]

7. **Solve the quadratic equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{33 \pm \sqrt{33^2 - 4 \cdot 1 \cdot 32}}{2 \cdot 1} \]
\[ x = \frac{33 \pm \sqrt{1089 - 128}}{2} \]
\[ x = \frac{33 \pm \sqrt{961}}{2} \]
\[ x = \frac{33 \pm 31}{2} \]
This gives us two solutions:
\[ x = \frac{64}{2} = 32 \quad \text{and} \quad x = \frac{2}{2} = 1 \]

8. **Back substitute to find \( r \)**:
Since \( x = r^5 \):
- For \( x = 32 \):
\[ r^5 = 32 \implies r = 2 \]
- For \( x = 1 \):
\[ r^5 = 1 \implies r = 1 \]

9. **Conclusion**:
The common ratio \( r \) can be either \( 2 \) or \( 1 \). However, since \( r = 1 \) does not form a G.P. (as all terms would be the same), we conclude:
\[ r = 2 \]

### Final Answer:
The common ratio \( r \) is \( 2 \).