If `a`, `b`, `c` are positive numbers such that `a gt b gt c` and the equation `(a+b-2c)x^(2)+(b+c-2a)x+(c+a-2b)=0` has a root in the interval `(-1,0)`, then

Options

A. `b` cannot be the `G.M.` of `a`,`c`
B. `b` may be the `G.M.` of `a`,`c`
C. `b` is the `G.M.` of `a`,`c`
D. none of these

Explanation

To solve the given problem, we need to analyze the quadratic equation:

\[ f(x) = (a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) \]

We want to find out if this equation has a root in the interval \((-1, 0)\). According to the Intermediate Value Theorem, for a quadratic equation to have a root in the interval \((-1, 0)\), the product of the function values at the endpoints of the interval must be negative:

\[ f(-1) \cdot f(0) < 0 \]

### Step 1: Calculate \(f(-1)\)

Substituting \(x = -1\) into the function:

\[ f(-1) = (a + b - 2c)(-1)^2 + (b + c - 2a)(-1) + (c + a - 2b) \]

This simplifies to:

\[ f(-1) = (a + b - 2c) - (b + c - 2a) + (c + a - 2b) \]

Combining like terms:

\[ f(-1) = a + b - 2c - b - c + 2a + c + a - 2b \]
\[ = 4a - 4b + 0 \]
\[ = 4(a - b) \]

### Step 2: Calculate \(f(0)\)

Substituting \(x = 0\) into the function:

\[ f(0) = (c + a - 2b) \]

### Step 3: Set up the inequality

Now we need to check the condition:

\[ f(-1) \cdot f(0) < 0 \]

Substituting the expressions we found:

\[ (4(a - b))(c + a - 2b) < 0 \]

### Step 4: Analyze the signs

Since \(a > b > c\), we know:

- \(a - b < 0\) (since \(a > b\))
- \(c + a - 2b\) needs to be analyzed further.

### Step 5: Analyze \(c + a - 2b\)

Since \(a > b > c\), we can conclude that \(c + a - 2b\) can be positive or negative depending on the specific values of \(a\), \(b\), and \(c\).

### Conclusion

To ensure that \(f(-1) \cdot f(0) < 0\), we need \(c + a < 2b\). Thus, the condition that must hold is:

\[ c + a < 2b \]

This condition ensures that the quadratic equation has a root in the interval \((-1, 0)\).