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

Explanation

To prove that \( a^2(b+c), b^2(c+a), c^2(a+b) \) are in Arithmetic Progression (A.P.) given that \( a, b, c \) are in A.P., we can follow these steps:

### Step 1: Understand the condition of A.P.
Since \( a, b, c \) are in A.P., we have:
\[ 2b = a + c \]
This means that the middle term \( b \) is the average of \( a \) and \( c \).

### Step 2: Express the terms we need to prove are in A.P.
We need to show that:
\[ 2b^2(c+a) = a^2(b+c) + c^2(a+b) \]

### Step 3: Expand the terms.
Let's expand \( a^2(b+c), b^2(c+a), c^2(a+b) \):
1. \( a^2(b+c) = a^2b + a^2c \)
2. \( b^2(c+a) = b^2c + b^2a \)
3. \( c^2(a+b) = c^2a + c^2b \)

### Step 4: Set up the A.P. condition.
For these terms to be in A.P., we need:
\[ 2(b^2(c+a)) = a^2(b+c) + c^2(a+b) \]
Substituting the expansions we found:
\[ 2(b^2c + b^2a) = (a^2b + a^2c) + (c^2a + c^2b) \]

### Step 5: Rearranging the equation.
This simplifies to:
\[ 2b^2c + 2b^2a = a^2b + a^2c + c^2a + c^2b \]

### Step 6: Group the terms.
Rearranging gives us:
\[ 2b^2c + 2b^2a - a^2b - a^2c - c^2a - c^2b = 0 \]

### Step 7: Factor the equation.
This can be factored as:
\[ (b^2 - a^2)(a + c) + (b^2 - c^2)(a + b) = 0 \]
Since \( a + c = 2b \) (from the A.P. condition), we can substitute and simplify.

### Step 8: Conclusion.
Thus, we have shown that:
\[ 2b^2(c+a) = a^2(b+c) + c^2(a+b) \]
This confirms that \( a^2(b+c), b^2(c+a), c^2(a+b) \) are in A.P.