If the sides of a triangle a,b,c be in A.P., then `a cos^2 ""C/2 +c cos^2 ""A/2 =`

Options

A. `(3a)/2`
B. `(3b)/2`
C. `(3c)/2`
D. `(Sigma a)/2`

Answer

To solve the problem, we need to show that if the sides of a triangle \( a, b, c \) are in Arithmetic Progression (A.P.), then the expression \( a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} \) simplifies to a certain value.

### Step-by-Step Solution:

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

2. **Using the Cosine Half-Angle Formula**:
The cosine half-angle formula states that:
\[ \cos^2 \frac{C}{2} = \frac{a}{2s} \quad \text{and} \quad \cos^2 \frac{A}{2} = \frac{c}{2s} \]
where \( s \) is the semi-perimeter of the triangle given by:
\[ s = \frac{a + b + c}{2} \]

3. **Substituting the Semi-Perimeter**:
From the A.P. condition, we can express \( s \) as:
\[ s = \frac{a + b + c}{2} = \frac{a + (a+c)/2 + c}{2} = \frac{2b + b}{2} = \frac{3b}{2} \]

4. **Substituting Values into the Expression**:
Now substituting \( s \) into the cosine half-angle formulas:
\[ \cos^2 \frac{C}{2} = \frac{a}{2 \cdot \frac{3b}{2}} = \frac{a}{3b} \]
\[ \cos^2 \frac{A}{2} = \frac{c}{2 \cdot \frac{3b}{2}} = \frac{c}{3b} \]

5. **Calculating the Required Expression**:
Now substituting these values into the expression \( a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} \):
\[ a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = a \cdot \frac{a}{3b} + c \cdot \frac{c}{3b} \]
\[ = \frac{a^2}{3b} + \frac{c^2}{3b} = \frac{a^2 + c^2}{3b} \]

6. **Using the A.P. Condition Again**:
From Equation 1, we know \( a + c = 2b \). Therefore, we can express \( a^2 + c^2 \) using the identity:
\[ a^2 + c^2 = (a + c)^2 - 2ac = (2b)^2 - 2ac = 4b^2 - 2ac \]

7. **Final Expression**:
Thus, substituting back, we have:
\[ a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{4b^2 - 2ac}{3b} \]

8. **Simplifying Further**:
Since \( a + c = 2b \), we can conclude that:
\[ a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{4b^2}{3b} - \frac{2ac}{3b} = \frac{4b}{3} - \frac{2ac}{3b} \]

### Conclusion:
Thus, the final result is:
\[ \boxed{\frac{4b}{3} - \frac{2ac}{3b}} \]