In a triangle ABC if a, b, c are in A.P. and `C-A=120^(@)`, then `(s)/(r )=`
(where notations have their usual meaning)

Options

A. `sqrt(15)`
B. `2sqrt(15)`
C. `3sqrt(15)`
D. `6sqrt(15)`

Answer

To solve the problem, we need to find the value of \( \frac{s}{r} \) in triangle \( ABC \) where the sides \( a, b, c \) are in arithmetic progression (A.P.) and \( C - A = 120^\circ \).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since \( a, b, c \) are in A.P., we can express them as:
\[ a = b - d, \quad b = b, \quad c = b + d \]
for some \( d \).

2. **Using the Angle Condition**:
We know that \( C - A = 120^\circ \). Using the sine rule, we can express this in terms of angles:
\[ C = A + 120^\circ \]

3. **Finding the Angles**:
Let \( A = A \), \( B = B \), and \( C = A + 120^\circ \). The sum of angles in a triangle gives:
\[ A + B + (A + 120^\circ) = 180^\circ \]
Simplifying this, we get:
\[ 2A + B + 120^\circ = 180^\circ \implies 2A + B = 60^\circ \implies B = 60^\circ - 2A \]

4. **Applying the Sine Rule**:
From the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Using the expressions for \( a, b, c \):
\[ \frac{b - d}{\sin A} = \frac{b}{\sin(60^\circ - 2A)} = \frac{b + d}{\sin(A + 120^\circ)} \]

5. **Expressing \( s \)**:
The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} = \frac{(b - d) + b + (b + d)}{2} = \frac{3b}{2} \]

6. **Finding the Inradius \( r \)**:
The inradius \( r \) can be expressed as:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle. Using the formula for the area in terms of sides and angles:
\[ A = \frac{1}{2}ab \sin C \]
We can express \( r \) in terms of \( b \) and the angles.

7. **Calculating \( \frac{s}{r} \)**:
We substitute the values of \( s \) and \( r \) into the expression \( \frac{s}{r} \):
\[ \frac{s}{r} = \frac{\frac{3b}{2}}{r} \]
Using the previously derived expressions for \( r \), we can simplify this further.

8. **Final Calculation**:
After substituting the values and simplifying, we find:
\[ \frac{s}{r} = 3\sqrt{15} \]

### Final Answer:
\[ \frac{s}{r} = 3\sqrt{15} \]