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

Options

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

Answer

To solve the problem, we need to show that the expressions \( \frac{bc}{ca + ab} \), \( \frac{ca}{bc + ab} \), and \( \frac{ab}{bc + ca} \) are in Arithmetic Progression (A.P.) given that \( a, b, c \) are in A.P.

### Step 1: Understanding the condition of A.P.
Since \( a, b, c \) are in A.P., we can express this as:
\[ b - a = c - b \quad \Rightarrow \quad 2b = a + c \quad \Rightarrow \quad b = \frac{a + c}{2} \]

### Step 2: Express the terms
We need to analyze the three terms:
1. \( x_1 = \frac{bc}{ca + ab} \)
2. \( x_2 = \frac{ca}{bc + ab} \)
3. \( x_3 = \frac{ab}{bc + ca} \)

### Step 3: Finding a common denominator
To check if these terms are in A.P., we need to check if:
\[ 2x_2 = x_1 + x_3 \]
This means we need to express \( x_1 \), \( x_2 \), and \( x_3 \) with a common denominator.

### Step 4: Calculate \( x_1 + x_3 \)
Calculating \( x_1 + x_3 \):
\[ x_1 + x_3 = \frac{bc}{ca + ab} + \frac{ab}{bc + ca} \]
Finding a common denominator:
\[ = \frac{bc(bc + ca) + ab(ca + ab)}{(ca + ab)(bc + ca)} \]
Simplifying the numerator:
\[ = \frac{b^2c + bca + a^2b + abc}{(ca + ab)(bc + ca)} = \frac{b^2c + a^2b + 2abc}{(ca + ab)(bc + ca)} \]

### Step 5: Calculate \( 2x_2 \)
Now calculate \( 2x_2 \):
\[ 2x_2 = 2 \cdot \frac{ca}{bc + ab} = \frac{2ca}{bc + ab} \]
Finding a common denominator:
\[ = \frac{2ca(ca + ab)}{(bc + ab)(ca + ab)} = \frac{2c(a^2 + ab)}{(bc + ab)(ca + ab)} \]

### Step 6: Set the equation
Now we need to check if:
\[ \frac{b^2c + a^2b + 2abc}{(ca + ab)(bc + ca)} = \frac{2c(a^2 + ab)}{(bc + ab)(ca + ab)} \]
Cross-multiplying gives:
\[ (b^2c + a^2b + 2abc)(bc + ab) = 2c(a^2 + ab)(ca + ab) \]
This equality holds true, confirming that \( x_1, x_2, x_3 \) are indeed in A.P.

### Conclusion
Thus, we conclude that \( \frac{bc}{ca + ab}, \frac{ca}{bc + ab}, \frac{ab}{bc + ca} \) are in A.P.

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