If the sides of a `Delta ABC` are in AP and a is the smallest sie, then cos A equals

Answer

To solve the problem, we need to find the value of cos A given that the sides of triangle ABC are in Arithmetic Progression (AP) and that side a is the smallest side.

### Step-by-Step Solution:

1. **Understanding the sides in AP**:
Since the sides of triangle ABC are in AP, we can denote the sides as:
- \( a \) (smallest side)
- \( b \) (middle side)
- \( c \) (largest side)

From the property of AP, we have:
\[ 2b = a + c \]

2. **Expressing \( a \) in terms of \( b \) and \( c \)**:
Rearranging the equation gives:
\[ a = 2b - c \]

3. **Using the Cosine Rule**:
The cosine of angle A can be expressed using the cosine rule:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]

4. **Substituting for \( a^2 \)**:
We need to find \( a^2 \). Substituting \( a = 2b - c \) into \( a^2 \):
\[ a^2 = (2b - c)^2 = 4b^2 - 4bc + c^2 \]

5. **Substituting \( a^2 \) back into the cosine formula**:
Now we substitute \( a^2 \) into the cosine formula:
\[ \cos A = \frac{b^2 + c^2 - (4b^2 - 4bc + c^2)}{2bc} \]

6. **Simplifying the expression**:
Simplifying the numerator:
\[ \cos A = \frac{b^2 + c^2 - 4b^2 + 4bc - c^2}{2bc} \]
This simplifies to:
\[ \cos A = \frac{4bc - 3b^2}{2bc} \]

7. **Final simplification**:
We can further simplify this to:
\[ \cos A = \frac{4c - 3b}{2c} \]

Thus, the final result for \( \cos A \) is:
\[ \cos A = \frac{4c - 3b}{2c} \]