In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125: 152.
Find its common ratio.

Answer

To find the common ratio of a geometric progression (G.P.) where the ratio between the sum of the first three terms and the sum of the first six terms is given as 125:152, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given the ratio of the sum of the first three terms (S3) to the sum of the first six terms (S6) of a G.P. as:
\[ \frac{S_3}{S_6} = \frac{125}{152} \]

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

3. **Applying the Formula**:
For the first three terms:
\[ S_3 = a \frac{r^3 - 1}{r - 1} \]
For the first six terms:
\[ S_6 = a \frac{r^6 - 1}{r - 1} \]

4. **Setting Up the Equation**:
Now, substituting these into the ratio:
\[ \frac{S_3}{S_6} = \frac{a \frac{r^3 - 1}{r - 1}}{a \frac{r^6 - 1}{r - 1}} = \frac{r^3 - 1}{r^6 - 1} \]
Since \( a \) and \( r - 1 \) cancel out, we have:
\[ \frac{r^3 - 1}{r^6 - 1} = \frac{125}{152} \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 152(r^3 - 1) = 125(r^6 - 1) \]
Expanding both sides:
\[ 152r^3 - 152 = 125r^6 - 125 \]

6. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ 125r^6 - 152r^3 + 27 = 0 \]

7. **Letting \( x = r^3 \)**:
Let \( x = r^3 \), then the equation becomes:
\[ 125x^2 - 152x + 27 = 0 \]

8. **Using the Quadratic Formula**:
We can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 125 \), \( b = -152 \), and \( c = 27 \):
\[ x = \frac{152 \pm \sqrt{(-152)^2 - 4 \cdot 125 \cdot 27}}{2 \cdot 125} \]

9. **Calculating the Discriminant**:
Calculate the discriminant:
\[ (-152)^2 = 23104, \quad 4 \cdot 125 \cdot 27 = 13500 \]
Thus,
\[ 23104 - 13500 = 9604 \]
So,
\[ \sqrt{9604} = 98 \]

10. **Finding the Roots**:
Now substituting back:
\[ x = \frac{152 \pm 98}{250} \]
This gives us two possible values for \( x \):
\[ x_1 = \frac{250}{250} = 1, \quad x_2 = \frac{54}{250} = \frac{27}{125} \]

11. **Finding the Common Ratio**:
Since \( x = r^3 \):
- For \( x_1 = 1 \): \( r = 1 \)
- For \( x_2 = \frac{27}{125} \): \( r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5} \)

12. **Conclusion**:
The common ratio \( r \) is:
\[ r = \frac{3}{5} \]

### Final Answer:
Hence, the common ratio is \( \frac{3}{5} \).