The second, third and sixth terms of an A.P. are consecutive terms of a geometric progression. Find the common ratio of the geometric progression.

Answer

To solve the problem, we need to find the common ratio of the geometric progression formed by the second, third, and sixth terms of an arithmetic progression (A.P.). Let's denote the first term of the A.P. as \( A \) and the common difference as \( D \).

### Step-by-Step Solution:

1. **Identify the Terms of the A.P.**:
- The second term of the A.P. is \( A_2 = A + D \).
- The third term of the A.P. is \( A_3 = A + 2D \).
- The sixth term of the A.P. is \( A_6 = A + 5D \).

2. **Set Up the Relationship for the GP**:
- According to the problem, \( A_2, A_3, A_6 \) are consecutive terms of a geometric progression (G.P.). Therefore, we can write:
\[ \frac{A_3}{A_2} = \frac{A_6}{A_3} \]
- Substituting the terms we found:
\[ \frac{A + 2D}{A + D} = \frac{A + 5D}{A + 2D} \]

3. **Cross Multiply**:
- Cross multiplying gives us:
\[ (A + 2D)^2 = (A + D)(A + 5D) \]

4. **Expand Both Sides**:
- Expanding the left side:
\[ (A + 2D)^2 = A^2 + 4AD + 4D^2 \]
- Expanding the right side:
\[ (A + D)(A + 5D) = A^2 + 5AD + AD + 5D^2 = A^2 + 6AD + 5D^2 \]

5. **Set the Equations Equal**:
- Now we have:
\[ A^2 + 4AD + 4D^2 = A^2 + 6AD + 5D^2 \]

6. **Simplify the Equation**:
- Cancel \( A^2 \) from both sides:
\[ 4AD + 4D^2 = 6AD + 5D^2 \]
- Rearranging gives:
\[ 4D^2 - 5D^2 = 6AD - 4AD \]
\[ -D^2 = 2AD \]

7. **Factor Out D**:
- Assuming \( D \neq 0 \) (since we need a valid A.P.), we can divide by \( D \):
\[ -D = 2A \]
- Thus, we have:
\[ D = -2A \]

8. **Find the Common Ratio**:
- Now we can find the common ratio \( r \) of the G.P.:
\[ r = \frac{A + 2D}{A + D} \]
- Substitute \( D = -2A \):
\[ r = \frac{A + 2(-2A)}{A + (-2A)} = \frac{A - 4A}{A - 2A} = \frac{-3A}{-A} = 3 \]

### Final Answer:
The common ratio of the geometric progression is \( \boxed{3} \).