If a, b, c are in A.P. as well as in G.P. then correct statement is :

Options

A. `a=b=c`
B. `anebnec`
C. `a=bnec`
D. None of these

Explanation

To solve the problem, we need to show that if \( a, b, c \) are in both Arithmetic Progression (A.P.) and Geometric Progression (G.P.), then they must be equal.

### Step-by-Step Solution:

1. **Understanding A.P. and G.P.**:
- For numbers \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
- For numbers \( a, b, c \) to be in G.P., the condition is:
\[ b^2 = ac \]

2. **Expressing in terms of G.P.**:
- Since \( a, b, c \) are in G.P., we can express \( b \) and \( c \) in terms of \( a \) and a common ratio \( r \):
\[ b = ar \quad \text{and} \quad c = ar^2 \]

3. **Substituting into A.P. condition**:
- Substitute \( b \) and \( c \) into the A.P. condition:
\[ 2(ar) = a + ar^2 \]
- Simplifying this gives:
\[ 2ar = a + ar^2 \]

4. **Rearranging the equation**:
- Rearranging the equation yields:
\[ 2ar - ar^2 - a = 0 \]
- Factoring out \( a \):
\[ a(2r - r^2 - 1) = 0 \]

5. **Analyzing the factors**:
- This gives us two cases:
1. \( a = 0 \) (which is trivial)
2. \( 2r - r^2 - 1 = 0 \)

6. **Solving the quadratic equation**:
- Rearranging the second case:
\[ r^2 - 2r + 1 = 0 \]
- This can be factored as:
\[ (r - 1)^2 = 0 \]
- Thus, we find:
\[ r = 1 \]

7. **Conclusion**:
- If \( r = 1 \), then:
\[ b = a \quad \text{and} \quad c = a \]
- Therefore, \( a = b = c \).

Thus, the correct statement is that if \( a, b, c \) are in both A.P. and G.P., then \( a = b = c \).