If `3x^(2)-2ax+(a^(2)+2b^(2)+2c^(2))=2(ab+bc)`, then `a`, `b`, `c` can be in

Options

A. `A.P.`
B. `G.P.`
C. `H.P.`
D. None of these

Explanation

To solve the equation \(3x^2 - 2ax + (a^2 + 2b^2 + 2c^2) = 2(ab + bc)\), we need to analyze the given polynomial and determine the relationship between \(a\), \(b\), and \(c\).

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start with the original equation:
\[ 3x^2 - 2ax + (a^2 + 2b^2 + 2c^2) - 2(ab + bc) = 0 \]
This simplifies to:
\[ 3x^2 - 2ax + (a^2 - 2ab + 2b^2 + 2c^2 - 2bc) = 0 \]

2. **Grouping Terms**:
We can rewrite the constant term:
\[ a^2 - 2ab + 2b^2 + 2c^2 - 2bc = (a - b)^2 + (b - c)^2 + (c - c)^2 \]
This means we can express the equation as:
\[ 3x^2 - 2ax + \left((a - b)^2 + (b - c)^2 + (c - c)^2\right) = 0 \]

3. **Identifying Roots**:
The equation can be expressed in terms of squares:
\[ (x - (a + b))^2 + (x - (b + c))^2 + (x - c)^2 = 0 \]
Since these are squares, each term must equal zero for the entire expression to equal zero.

4. **Setting Each Square to Zero**:
From the squares, we have:
\[ x - (a + b) = 0 \quad \Rightarrow \quad x = a + b \]
\[ x - (b + c) = 0 \quad \Rightarrow \quad x = b + c \]
\[ x - c = 0 \quad \Rightarrow \quad x = c \]

5. **Equating the Roots**:
Since all expressions equal \(x\), we can equate them:
\[ a + b = b + c = c \]

6. **Finding Relationships**:
From \(a + b = b + c\), we can simplify to:
\[ a = c \]
From \(b + c = c\), we can simplify to:
\[ b = 0 \]

7. **Conclusion**:
Since we have \(a = c\) and \(b = 0\), we can express \(a\), \(b\), and \(c\) as:
\[ a, 0, a \]
This indicates that \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP) because the middle term \(b\) is the average of \(a\) and \(c\).

### Final Answer:
Thus, \(a\), \(b\), and \(c\) can be in **Arithmetic Progression (AP)**.