The second, the first and the third term of an A.P whose common difference is non zero, form a G.P. in that order. Find its common ratio.

Options

A. `-1`
B. 1
C. `-2`
D. 2

Answer

To solve the problem, we need to find the common ratio of a geometric progression (G.P.) formed by the second, first, and third terms of an arithmetic progression (A.P.) where the common difference is non-zero.

### Step-by-Step Solution:

1. **Define the terms of the A.P.**:
Let the first term of the A.P. be \( A \) and the common difference be \( D \). The terms of the A.P. can be expressed as:
- First term: \( A \)
- Second term: \( A + D \)
- Third term: \( A + 2D \)

2. **Identify the order of terms forming the G.P.**:
According to the problem, the second, first, and third terms of the A.P. form a G.P. in that order. Thus, we have:
- First term of G.P.: \( A + D \)
- Second term of G.P.: \( A \)
- Third term of G.P.: \( A + 2D \)

3. **Set up the relationship for G.P.**:
For three terms to be in G.P., the square of the middle term must equal the product of the other two terms. Therefore, we can write:
\[ (A)^2 = (A + D)(A + 2D) \]

4. **Expand the equation**:
Expanding the right-hand side:
\[ A^2 = A^2 + 2AD + AD + 2D^2 \]
Simplifying gives:
\[ A^2 = A^2 + 3AD + 2D^2 \]

5. **Rearranging the equation**:
Subtract \( A^2 \) from both sides:
\[ 0 = 3AD + 2D^2 \]

6. **Factoring the equation**:
Factor out \( D \):
\[ D(3A + 2D) = 0 \]

7. **Solving for D**:
Since the common difference \( D \) is non-zero, we can set:
\[ 3A + 2D = 0 \]
This implies:
\[ 2D = -3A \quad \Rightarrow \quad D = -\frac{3}{2}A \]

8. **Finding the common ratio**:
The common ratio \( R \) of the G.P. can be calculated as:
\[ R = \frac{\text{Second term}}{\text{First term}} = \frac{A}{A + D} \]
Substituting \( D = -\frac{3}{2}A \):
\[ R = \frac{A}{A - \frac{3}{2}A} = \frac{A}{-\frac{1}{2}A} = -2 \]

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