In a `DeltaABC, bcos^(2)'(A)/(2) + acos^(2)"(B)/(2) = (3)/(2)c`, then `a, c, b` in (with usual notations)

Options

A. `A.P`
B. `G.P`
C. `H.P`
D. `A.G.P.`

Explanation

To solve the problem step by step, we start with the given equation:

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

### Step 1: Use the formula for \(\cos^2 \left(\frac{A}{2}\right)\) and \(\cos^2 \left(\frac{B}{2}\right)\)

We know that:

\[ \cos^2 \left(\frac{A}{2}\right) = \frac{s(s-a)}{bc} \]
\[ \cos^2 \left(\frac{B}{2}\right) = \frac{s(s-b)}{ac} \]

where \(s\) is the semi-perimeter given by \(s = \frac{a+b+c}{2}\).

### Step 2: Substitute these formulas into the equation

Substituting the formulas into the original equation gives:

\[ \frac{b \cdot \frac{s(s-a)}{bc}}{2} + \frac{a \cdot \frac{s(s-b)}{ac}}{2} = \frac{3}{2} c \]

This simplifies to:

\[ \frac{s(s-a)}{2c} + \frac{s(s-b)}{2c} = \frac{3}{2} c \]

### Step 3: Combine the terms on the left side

Combining the left side:

\[ \frac{s(s-a+s-b)}{2c} = \frac{3}{2} c \]

This simplifies to:

\[ \frac{s(2s - a - b)}{2c} = \frac{3}{2} c \]

### Step 4: Cross-multiply to eliminate the fraction

Cross-multiplying gives:

\[ s(2s - a - b) = 3c^2 \]

### Step 5: Substitute \(s\) with \(\frac{a+b+c}{2}\)

Now substituting \(s = \frac{a+b+c}{2}\):

\[ \frac{a+b+c}{2}(2\cdot\frac{a+b+c}{2} - a - b) = 3c^2 \]

This simplifies to:

\[ \frac{a+b+c}{2}(c) = 3c^2 \]

### Step 6: Simplify the equation

This leads to:

\[ (a+b+c)c = 6c^2 \]

Assuming \(c \neq 0\), we can divide both sides by \(c\):

\[ a + b + c = 6c \]

### Step 7: Rearranging the equation

Rearranging gives:

\[ a + b = 5c \]

### Step 8: Conclude the relationship

This implies that \(a, b, c\) are in arithmetic progression since:

1. \(a + b = 5c\)
2. Thus, \(a + b + c = 6c\)

So, we can express \(a\) and \(b\) in terms of \(c\):

- Let \(a = 2c\)
- Let \(b = 3c\)

Thus, \(a, b, c\) are in arithmetic progression.

### Final Result

The relationship among \(a, b, c\) is:

\[ a + b = 5c \quad \text{(in AP)} \]

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