The sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. The common ratio of the G.P. is

Options

A. `1//5`
B. `2//5`
C. `3//5`
D. `4//5`

Answer

To solve the problem, we need to find the common ratio of a geometric progression (G.P.) given that the ratio of the sum of the first three terms to the sum of the first six terms is 125:152.

### Step-by-Step Solution:

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

2. **Calculate the sums for \( n = 3 \) and \( n = 6 \)**:
- For the first three terms:
\[ S_3 = A \frac{r^3 - 1}{r - 1} \]
- For the first six terms:
\[ S_6 = A \frac{r^6 - 1}{r - 1} \]

3. **Set up the ratio of the sums**:
According to the problem, the ratio of \( S_3 \) to \( S_6 \) is given as:
\[ \frac{S_3}{S_6} = \frac{125}{152} \]
Substituting the formulas for \( S_3 \) and \( S_6 \):
\[ \frac{A \frac{r^3 - 1}{r - 1}}{A \frac{r^6 - 1}{r - 1}} = \frac{125}{152} \]
The \( A \) and \( (r - 1) \) terms cancel out:
\[ \frac{r^3 - 1}{r^6 - 1} = \frac{125}{152} \]

4. **Simplify the equation**:
We can rewrite \( r^6 - 1 \) using the difference of squares:
\[ r^6 - 1 = (r^3 - 1)(r^3 + 1) \]
Thus, we can rewrite the ratio:
\[ \frac{r^3 - 1}{(r^3 - 1)(r^3 + 1)} = \frac{125}{152} \]
This simplifies to:
\[ \frac{1}{r^3 + 1} = \frac{125}{152} \]

5. **Cross-multiply to solve for \( r^3 \)**:
Cross-multiplying gives:
\[ 152 = 125(r^3 + 1) \]
Expanding this:
\[ 152 = 125r^3 + 125 \]
Rearranging gives:
\[ 125r^3 = 152 - 125 \]
\[ 125r^3 = 27 \]

6. **Solve for \( r^3 \)**:
Dividing both sides by 125:
\[ r^3 = \frac{27}{125} \]

7. **Find the common ratio \( r \)**:
Taking the cube root of both sides:
\[ r = \sqrt[3]{\frac{27}{125}} = \frac{3}{5} \]

### Final Answer:
The common ratio \( r \) of the G.P. is \( \frac{3}{5} \).

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