If ` a^2+2bc ,b^2+2ca, c^2+2ab` are in A.P. then :-

Options

A. `(a-b)(c-a),(a-b)(b-c),(b-c)(c-a)` are in A.P
B. b-c,c-a,a-b are in H.P
C. a+b,b+c,c+a are in H.P
D. `a^2,b^2,c^2` are in H.P

Explanation

To solve the problem, we need to show that if \( a^2 + 2bc, b^2 + 2ca, c^2 + 2ab \) are in Arithmetic Progression (A.P.), then we can derive certain relationships among \( a, b, \) and \( c \).

### Step-by-Step Solution:

1. **Understanding the Condition for A.P.**:
Since \( a^2 + 2bc, b^2 + 2ca, c^2 + 2ab \) are in A.P., the condition for three terms \( x, y, z \) to be in A.P. is:
\[ 2y = x + z \]
Applying this to our terms, we have:
\[ 2(b^2 + 2ca) = (a^2 + 2bc) + (c^2 + 2ab) \]

2. **Expanding the Equation**:
Expanding both sides gives:
\[ 2b^2 + 4ca = a^2 + c^2 + 2bc + 2ab \]

3. **Rearranging the Terms**:
Rearranging the equation leads to:
\[ 2b^2 - 2bc - 2ab + 4ca - a^2 - c^2 = 0 \]
Simplifying this, we can factor out 2:
\[ 2(b^2 - bc - ab + 2ca) - (a^2 + c^2) = 0 \]

4. **Grouping Terms**:
We can rearrange the terms further:
\[ b^2 - ab - bc + 2ca = \frac{1}{2}(a^2 + c^2) \]

5. **Analyzing the Result**:
The equation \( b^2 - ab - bc + 2ca = 0 \) can be analyzed to find relationships among \( a, b, \) and \( c \).

6. **Conclusion**:
From the derived relationships, we can conclude that the differences \( (b - a), (c - b), (a - c) \) are in A.P. or that the terms \( (c - a), (a - b), (b - c) \) are in Harmonic Progression (H.P.).

### Final Result:
Thus, the conclusion is that if \( a^2 + 2bc, b^2 + 2ca, c^2 + 2ab \) are in A.P., then the differences \( (b - a), (c - b), (a - c) \) are in A.P. or equivalently, \( (c - a), (a - b), (b - c) \) are in H.P.