If in a triangle `A B C ,acos^2(C/2)+ c cos^2(A/2)=(3b)/2,` then the sides `a ,b ,a n dc` `` are in A.P. b. are in G.P. c. are in H.P. d. satisfy `a+b=cdot`

Options

A. are in G.P.
B. are in H.P.
C. satisfy a+b+c
D. are in A.P.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the given equation
We start with the equation given in the problem:
\[ a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right) = \frac{3b}{2} \]

### Step 2: Use the half-angle formulas
We know the half-angle formulas for cosine in terms of the semi-perimeter \(s\):
- \(\cos^2\left(\frac{C}{2}\right) = \frac{s(s - c)}{ab}\)
- \(\cos^2\left(\frac{A}{2}\right) = \frac{s(s - a)}{bc}\)

### Step 3: Substitute the half-angle formulas into the equation
Substituting these formulas into the original equation, we have:
\[ a \cdot \frac{s(s - c)}{ab} + c \cdot \frac{s(s - a)}{bc} = \frac{3b}{2} \]

### Step 4: Simplify the equation
This simplifies to:
\[ \frac{s(s - c)}{b} + \frac{s(s - a)}{b} = \frac{3b}{2} \]
Combining the terms gives:
\[ \frac{s((s - c) + (s - a))}{b} = \frac{3b}{2} \]

### Step 5: Factor out \(s\)
Factoring out \(s\) from the left side:
\[ \frac{s(2s - a - c)}{b} = \frac{3b}{2} \]

### Step 6: Cross-multiply to eliminate the fraction
Cross-multiplying yields:
\[ 2s(2s - a - c) = 3b^2 \]

### Step 7: Substitute \(2s\) with \(a + b + c\)
Since \(2s = a + b + c\), we substitute this into the equation:
\[ (a + b + c)(2s - a - c) = 3b^2 \]

### Step 8: Simplify and rearrange
This leads to:
\[ (a + b + c)(b) = 3b^2 \]
Dividing both sides by \(b\) (assuming \(b \neq 0\)):
\[ a + b + c = 3b \]
Rearranging gives:
\[ a + c = 2b \]

### Step 9: Conclusion
Since \(a + c = 2b\), we can conclude that the sides \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).

### Final Answer
The sides \(a\), \(b\), and \(c\) are in AP.