If a,b,c are in A.P. then the roots of the equation `(a+b-c)x^2 + (b-a) x-a=0` are :

Options

A. 1,a
B. `-1,a`
C. 1,b
D. None of these

Answer

To solve the problem, we need to find the roots of the equation given that \( a, b, c \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understand the Condition for A.P.**:
Since \( a, b, c \) are in A.P., we have the relation:
\[ 2b = a + c \quad \text{(1)} \]

2. **Rewrite the Given Equation**:
The equation provided is:
\[ (a + b - c)x^2 + (b - a)x - a = 0 \]

3. **Substitute for \( c \)**:
From equation (1), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]
Now substitute \( c \) into the equation:
\[ (a + b - (2b - a))x^2 + (b - a)x - a = 0 \]
Simplifying this gives:
\[ (a + b - 2b + a)x^2 + (b - a)x - a = 0 \]
\[ (2a - b)x^2 + (b - a)x - a = 0 \]

4. **Identify Coefficients**:
The coefficients of the quadratic equation are:
- \( A = 2a - b \)
- \( B = b - a \)
- \( C = -a \)

5. **Find One Root**:
To find one of the roots, we can substitute \( x = 1 \) into the equation:
\[ (2a - b)(1)^2 + (b - a)(1) - a = 0 \]
This simplifies to:
\[ 2a - b + b - a - a = 0 \]
\[ 2a - a - a = 0 \]
Thus, \( x = 1 \) is a root.

6. **Find the Other Root Using the Sum of Roots**:
The sum of the roots \( \alpha + \beta \) of the quadratic equation is given by:
\[ \alpha + \beta = -\frac{B}{A} = -\frac{b - a}{2a - b} \]
Since one root \( \alpha = 1 \), we have:
\[ 1 + \beta = -\frac{b - a}{2a - b} \]
Therefore:
\[ \beta = -\frac{b - a}{2a - b} - 1 \]
Simplifying this gives:
\[ \beta = -\frac{b - a + (2a - b)}{2a - b} = -\frac{a}{2a - b} \]

7. **Final Roots**:
The roots of the equation are:
\[ x_1 = 1 \quad \text{and} \quad x_2 = -\frac{a}{2a - b} \]

### Summary of the Roots:
The roots of the equation are:
- \( x_1 = 1 \)
- \( x_2 = -\frac{a}{2a - b} \)