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

Explanation

To solve the problem, we need to find the common ratio \( r \) of a geometric progression (G.P.) given the ratio of the sum of the first three terms to the sum of the first six terms.

### Step 1: Write the formula for the sum of the first \( n \) terms of a G.P.
The sum of the first \( n \) terms \( S_n \) of a G.P. can be expressed as:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

### Step 2: Write the expressions for \( S_3 \) and \( S_6 \)
Using the formula, we can write:
\[ S_3 = \frac{a(1 - r^3)}{1 - r} \]
\[ S_6 = \frac{a(1 - r^6)}{1 - r} \]

### Step 3: Set up the ratio of \( S_3 \) to \( S_6 \)
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 expressions for \( S_3 \) and \( S_6 \):
\[ \frac{\frac{a(1 - r^3)}{1 - r}}{\frac{a(1 - r^6)}{1 - r}} = \frac{125}{152} \]
The \( a \) and \( (1 - r) \) terms cancel out (assuming \( r \neq 1 \)):
\[ \frac{1 - r^3}{1 - r^6} = \frac{125}{152} \]

### Step 4: Simplify the expression
We can express \( 1 - r^6 \) in terms of \( 1 - r^3 \):
\[ 1 - r^6 = (1 - r^3)(1 + r^3) \]
Substituting this into the ratio gives:
\[ \frac{1 - r^3}{(1 - r^3)(1 + r^3)} = \frac{125}{152} \]
This simplifies to:
\[ \frac{1}{1 + r^3} = \frac{125}{152} \]

### Step 5: Cross-multiply to solve for \( r^3 \)
Cross-multiplying gives:
\[ 152 = 125(1 + r^3) \]
Expanding this:
\[ 152 = 125 + 125r^3 \]
Subtracting 125 from both sides:
\[ 27 = 125r^3 \]
Dividing both sides by 125:
\[ r^3 = \frac{27}{125} \]

### Step 6: Solve for \( r \)
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:
\[ \boxed{\frac{3}{5}} \]