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.) when \( a, b, c \) are in A.P., we 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 \( c \).

**Hint:** Recall that in an A.P., the middle term is the average of the other two terms.

### Step 2: Write 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 expressions.
Let's expand each term:
1. \( a^2(b+c) = a^2b + a^2c \)
2. \( b^2(c+a) = b^2c + b^2a \)
3. \( c^2(a+b) = c^2a + c^2b \)

Now, we can write:
\[ a^2(b+c) + c^2(a+b) = a^2b + a^2c + c^2a + c^2b \]

### Step 4: Combine the terms.
Now, we need to combine the terms:
\[ a^2b + a^2c + c^2a + c^2b \]

### Step 5: Rearranging the equation.
We can rearrange the terms to group them:
\[ = a^2b + c^2b + a^2c + c^2a \]

### Step 6: Use the A.P. condition.
Using the condition \( 2b = a + c \), we can rewrite \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]

### Step 7: Substitute and simplify.
Substituting \( c \) into our expressions will help us simplify:
1. Substitute \( c \) into \( a^2(b+c) \) and \( c^2(a+b) \).
2. After substitution, simplify the expressions to show that they are equal to \( 2b^2(c+a) \).

### Step 8: Conclude the proof.
After simplification, if we find that:
\[ 2b^2(c+a) = a^2(b+c) + c^2(a+b) \]
Then we can conclude that \( a^2(b+c), b^2(c+a), c^2(a+b) \) are indeed in A.P.

**Final Statement:** Thus, we have proved that if \( a, b, c \) are in A.P., then \( a^2(b+c), b^2(c+a), c^2(a+b) \) are also in A.P.

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