If `a , b` and `c` are in `A P ,` then the straight line `a x+b y+c=0` will always pass through a fixed point whose coordinates are______

Explanation

To solve the problem, we need to find the fixed point through which the line \( ax + by + c = 0 \) passes when \( a, b, c \) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding Arithmetic Progression (AP)**:
- If \( a, b, c \) are in AP, then the condition for AP is given by:
\[ 2b = a + c \]
- This can be rearranged to:
\[ a - 2b + c = 0 \]

2. **Equation of the Line**:
- The equation of the line is given as:
\[ ax + by + c = 0 \]
- We can rewrite this equation in a form that highlights the relationship with the coefficients \( a, b, c \):
\[ ax + by = -c \]

3. **Substituting the AP Condition**:
- From the AP condition \( a - 2b + c = 0 \), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]

4. **Finding the Fixed Point**:
- Substitute \( c \) into the line equation:
\[ ax + by + (2b - a) = 0 \]
- Simplifying this gives:
\[ ax + by + 2b - a = 0 \]
- Rearranging yields:
\[ ax + by = a - 2b \]

5. **Finding the Coordinates**:
- To find the fixed point, we can choose specific values for \( x \) and \( y \) that satisfy the line equation regardless of the values of \( a \) and \( b \).
- Setting \( x = 1 \) and \( y = -2 \):
\[ a(1) + b(-2) + c = 0 \]
- This simplifies to:
\[ a - 2b + c = 0 \]
- Since we know from the AP condition that \( a - 2b + c = 0 \), this confirms that the point \( (1, -2) \) lies on the line.

### Conclusion:
The straight line \( ax + by + c = 0 \) will always pass through the fixed point whose coordinates are:
\[ (1, -2) \]