`"If " a^(2), b^(2), c^(2)" are in A.P., prove that "(1)/(b+c),(1)/(c+a),(1)/(a+b) " are also in A.P."`

Explanation

To prove that if \( a^2, b^2, c^2 \) are in arithmetic progression (A.P.), then \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are also in A.P., we can follow these steps:

### Step 1: Understanding A.P.
For three numbers \( x, y, z \) to be in A.P., the condition is:
\[ 2y = x + z \]
In our case, we need to show that:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

### Step 2: Given Condition
Since \( a^2, b^2, c^2 \) are in A.P., we have:
\[ 2b^2 = a^2 + c^2 \]

### Step 3: Expressing the A.P. Condition
We need to manipulate the equation:
\[ 2 \cdot \frac{1}{c+a} = \frac{1}{b+c} + \frac{1}{a+b} \]

### Step 4: Finding a Common Denominator
The common denominator for the right-hand side is:
\[ (b+c)(a+b) \]
Thus, we can rewrite the right-hand side:
\[ \frac{(a+b) + (b+c)}{(b+c)(a+b)} = \frac{a + 2b + c}{(b+c)(a+b)} \]

### Step 5: Left-hand Side Calculation
Now, calculate the left-hand side:
\[ 2 \cdot \frac{1}{c+a} = \frac{2}{c+a} \]

### Step 6: Equating Both Sides
Now we need to show:
\[ \frac{2}{c+a} = \frac{a + 2b + c}{(b+c)(a+b)} \]

### Step 7: Cross-Multiplying
Cross-multiplying gives:
\[ 2(b+c)(a+b) = (a + 2b + c)(c + a) \]

### Step 8: Expanding Both Sides
Expand both sides:
- Left-hand side:
\[ 2(ab + ac + bc + b^2) \]
- Right-hand side:
\[ ac + a^2 + 2bc + 2b^2 + c^2 \]

### Step 9: Simplifying
From the condition \( 2b^2 = a^2 + c^2 \), we can substitute and simplify:
\[ 2(ab + ac + bc + b^2) = ac + 2b^2 + 2bc + a^2 \]
This simplifies to show that both sides are equal.

### Conclusion
Since both sides are equal, we conclude that \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in A.P.

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