If a,b,c are in A.P. as well as is G.P. then the value of `a^(b-c)+b^(c-a)+c^(a-b)` is
(i) 1
(ii) 3
(iii) 6
(iv) None of these

Options

A. 1
B. 3
C. 6
D. None of these

Answer

To solve the problem, we need to analyze the conditions given: \( a, b, c \) are in both Arithmetic Progression (A.P.) and Geometric Progression (G.P.).

### Step-by-Step Solution:

1. **Understanding A.P. and G.P. Conditions:**
- Since \( a, b, c \) are in A.P., we have:
\[ 2b = a + c \quad \text{(Equation 1)} \]
- Since \( a, b, c \) are also in G.P., we have:
\[ b^2 = ac \quad \text{(Equation 2)} \]

2. **Using the A.P. Condition:**
- From Equation 1, we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]

3. **Substituting into the G.P. Condition:**
- Substitute \( c \) into Equation 2:
\[ b^2 = a(2b - a) \]
- Expanding this gives:
\[ b^2 = 2ab - a^2 \]
- Rearranging leads to:
\[ a^2 - 2ab + b^2 = 0 \]
- This can be factored as:
\[ (a - b)^2 = 0 \]
- Therefore, we conclude:
\[ a = b \]

4. **Finding \( c \):**
- Now substituting \( a = b \) into the A.P. condition:
\[ 2b = a + c \implies 2a = a + c \implies c = a \]
- Thus, we have:
\[ a = b = c \]

5. **Calculating the Expression:**
- We need to find the value of:
\[ a^{(b-c)} + b^{(c-a)} + c^{(a-b)} \]
- Since \( a = b = c \), we can substitute:
\[ a^{(b-c)} = a^{(a-a)} = a^0 = 1 \]
\[ b^{(c-a)} = b^{(a-a)} = b^0 = 1 \]
\[ c^{(a-b)} = c^{(a-a)} = c^0 = 1 \]
- Therefore, the entire expression becomes:
\[ 1 + 1 + 1 = 3 \]

6. **Final Answer:**
- The value of \( a^{(b-c)} + b^{(c-a)} + c^{(a-b)} \) is:
\[ \boxed{3} \]