If `a,b,c` are in A.P then `a+1/(bc), b+1/(ca), c+1/(ab)` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to determine if the expressions \( a + \frac{1}{bc}, b + \frac{1}{ca}, c + \frac{1}{ab} \) are in arithmetic progression (A.P.) given that \( a, b, c \) are in A.P.

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( a, b, c \) are in A.P., we know that:
\[ b - a = c - b \]
This implies:
\[ 2b = a + c \quad \text{or} \quad b = \frac{a + c}{2} \]

2. **Analyzing the New Expressions**:
We need to check if the following three terms are in A.P.:
\[ A_1 = a + \frac{1}{bc}, \quad A_2 = b + \frac{1}{ca}, \quad A_3 = c + \frac{1}{ab} \]

3. **Finding the Common Difference**:
For the three terms \( A_1, A_2, A_3 \) to be in A.P., the condition that must hold is:
\[ 2A_2 = A_1 + A_3 \]
Substituting the expressions, we get:
\[ 2\left(b + \frac{1}{ca}\right) = \left(a + \frac{1}{bc}\right) + \left(c + \frac{1}{ab}\right) \]

4. **Expanding the Equation**:
Expanding both sides:
\[ 2b + \frac{2}{ca} = a + c + \frac{1}{bc} + \frac{1}{ab} \]

5. **Rearranging Terms**:
Rearranging gives:
\[ 2b - (a + c) = \frac{1}{bc} + \frac{1}{ab} - \frac{2}{ca} \]
Since \( a + c = 2b \) (from the A.P. condition), the left side becomes:
\[ 2b - 2b = 0 \]
Therefore, we need to check if the right side also equals zero:
\[ \frac{1}{bc} + \frac{1}{ab} - \frac{2}{ca} = 0 \]

6. **Finding a Common Denominator**:
The common denominator for the right side is \( abc \):
\[ \frac{a + c - 2b}{abc} = 0 \]
Since \( a + c = 2b \), this simplifies to:
\[ \frac{0}{abc} = 0 \]

7. **Conclusion**:
Since both sides are equal, we conclude that:
\[ a + \frac{1}{bc}, b + \frac{1}{ca}, c + \frac{1}{ab} \text{ are in A.P.} \]

### Final Answer:
The expressions \( a + \frac{1}{bc}, b + \frac{1}{ca}, c + \frac{1}{ab} \) are in **Arithmetic Progression (A.P.)**.