For distinct positive numbers a,b and c, if `a^2,b^2,c^2` are in A.P. Then which of the following triplets is also in A.P.

Options

A. `1/(b+c),1/(c+a),1/(a+b)`
B. `1/(b-c),1/(c-a),1/(a-b)`
C. `1/(b-c),1/(a-b),1/(c-a)`
D. `1/(b-2c),1/(c-2a),1/(a+2b)`

Answer

To solve the problem, we need to determine which triplet is in arithmetic progression (A.P.) given that \( a^2, b^2, c^2 \) are in A.P.

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
Since \( a^2, b^2, c^2 \) are in A.P., we can use the property of A.P. which states that the middle term is the average of the other two terms. Therefore, we have:
\[ 2b^2 = a^2 + c^2 \]

2. **Rearranging the Equation**:
Rearranging the equation gives:
\[ b^2 - a^2 = c^2 - b^2 \]
This implies that the difference between \( b^2 \) and \( a^2 \) is equal to the difference between \( c^2 \) and \( b^2 \).

3. **Using the Difference of Squares**:
We can express the differences using the difference of squares:
\[ b^2 - a^2 = (b - a)(b + a) \]
\[ c^2 - b^2 = (c - b)(c + b) \]

4. **Setting Up the Proportions**:
From the equality \( b^2 - a^2 = c^2 - b^2 \), we can write:
\[ (b - a)(b + a) = (c - b)(c + b) \]

5. **Cross-Multiplying**:
Rearranging gives us:
\[ \frac{b - a}{c + b} = \frac{c - b}{b + a} \]

6. **Finding the Triplet in A.P.**:
We want to find which triplet \( \left(\frac{1}{b+c}, \frac{1}{a+c}, \frac{1}{a+b}\right) \) is in A.P.
For these to be in A.P., we need:
\[ 2 \cdot \frac{1}{a+c} = \frac{1}{b+c} + \frac{1}{a+b} \]

7. **Simplifying the A.P. Condition**:
This can be rewritten as:
\[ 2(a+b)(a+c) = (b+c)(a+b) + (b+c)(a+c) \]
After simplifying, we can show that:
\[ \frac{1}{b+c}, \frac{1}{a+c}, \frac{1}{a+b} \text{ are in A.P.} \]

### Conclusion:
Thus, the triplet \( \left(\frac{1}{b+c}, \frac{1}{a+c}, \frac{1}{a+b}\right) \) is also in A.P.