If `(b+c -a)/(a) , (c+a-b)/(b) , (a + b -c)/(c )` are in AP then which of the following is in AP ?

Options

A. a,b,c
B. `a^2, b^2, c^2`
C. `1/a, 1/b, 1/c`
D. none of these

Answer

To solve the problem, we need to determine if the expressions \((b+c-a)/a\), \((c+a-b)/b\), and \((a+b-c)/c\) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the Condition for AP**:
For three terms \(x\), \(y\), and \(z\) to be in AP, the condition is:
\[ 2y = x + z \]
Here, let:
\[ x = \frac{b+c-a}{a}, \quad y = \frac{c+a-b}{b}, \quad z = \frac{a+b-c}{c} \]

2. **Setting Up the Equation**:
We need to check if:
\[ 2 \left(\frac{c+a-b}{b}\right) = \frac{b+c-a}{a} + \frac{a+b-c}{c} \]

3. **Finding a Common Denominator**:
The common denominator for the right-hand side is \(abc\):
\[ \frac{b+c-a}{a} = \frac{(b+c-a)bc}{abc}, \quad \frac{a+b-c}{c} = \frac{(a+b-c)ab}{abc} \]
Thus, we can rewrite the right-hand side:
\[ \frac{(b+c-a)bc + (a+b-c)ab}{abc} \]

4. **Simplifying the Right-Hand Side**:
Expanding the numerator:
\[ (b+c-a)bc + (a+b-c)ab = b^2c + bc^2 - abc + a^2b + ab^2 - abc \]
Combine like terms:
\[ = b^2c + bc^2 + a^2b + ab^2 - 2abc \]

5. **Simplifying the Left-Hand Side**:
The left-hand side becomes:
\[ 2 \left(\frac{c+a-b}{b}\right) = \frac{2(c+a-b)a}{b} \]

6. **Equating Both Sides**:
Now we need to equate both sides:
\[ \frac{2(c+a-b)a}{b} = \frac{b^2c + bc^2 + a^2b + ab^2 - 2abc}{abc} \]

7. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 2(c+a-b)abc = b(b^2c + bc^2 + a^2b + ab^2 - 2abc) \]

8. **Identifying the Result**:
After simplifying, we find that the condition holds true if:
\[ \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text{ are in AP.} \]

### Conclusion:
Thus, if \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), and \(\frac{a+b-c}{c}\) are in AP, then \(\frac{1}{a}\), \(\frac{1}{b}\), and \(\frac{1}{c}\) are also in AP.