If in a triangles `a cos^(2)(C/2)+c cos^(2)(A/2)=(3b)/2`, then the sides of the triangle are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the given equation in the context of triangle properties. The equation provided is:

\[ a \cos^2\left(\frac{C}{2}\right) + c \cos^2\left(\frac{A}{2}\right) = \frac{3b}{2} \]

We will break down the solution step by step.

### Step 1: Use the identity for cosine
We know that:

\[ 1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right) \]

Using this identity, we can express \(\cos^2\left(\frac{C}{2}\right)\) and \(\cos^2\left(\frac{A}{2}\right)\):

\[ \cos^2\left(\frac{C}{2}\right) = \frac{1 + \cos C}{2} \]
\[ \cos^2\left(\frac{A}{2}\right) = \frac{1 + \cos A}{2} \]

### Step 2: Substitute into the original equation
Substituting these into the original equation gives:

\[ a \left(\frac{1 + \cos C}{2}\right) + c \left(\frac{1 + \cos A}{2}\right) = \frac{3b}{2} \]

Multiplying through by 2 to eliminate the fractions:

\[ a(1 + \cos C) + c(1 + \cos A) = 3b \]

### Step 3: Expand and rearrange
Expanding this, we get:

\[ a + a \cos C + c + c \cos A = 3b \]

Rearranging gives:

\[ a \cos C + c \cos A = 3b - (a + c) \]

### Step 4: Use the Law of Cosines
Using the Law of Cosines, we have:

\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 5: Substitute the cosine values
Substituting these values into the rearranged equation:

\[ a \left(\frac{a^2 + b^2 - c^2}{2ab}\right) + c \left(\frac{b^2 + c^2 - a^2}{2bc}\right) = 3b - (a + c) \]

### Step 6: Simplify and analyze
This equation can be simplified further, but we notice that if we analyze the condition \(A + C = 2B\), it implies that the sides \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).

### Conclusion
Thus, from the analysis, we conclude that the sides of the triangle \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).