Sides of triangle ABC, a, b, c are in G.P. If 'r' be the common ratio of this G.P. , then

Options

A. `r in ((sqrt5-1)/(2), oo)`
B. `r in ((sqrt5+1)/(2), oo)`
C. `r in ((sqrt5-1)/(2),(sqrt5+1)/(2))`
D. `r in ((sqrt5+1)/(2), (sqrt5+3)/(2))`

Answer

To solve the problem where the sides of triangle ABC, denoted as \( a, b, c \), are in a geometric progression (G.P.) with a common ratio \( r \), we can follow these steps:

### Step 1: Define the sides of the triangle in terms of the common ratio
Since the sides are in G.P., we can express them as:
- \( a = a_1 \)
- \( b = a_1 r \)
- \( c = a_1 r^2 \)

### Step 2: Apply the triangle inequality conditions
For any triangle with sides \( a, b, c \), the following inequalities must hold:
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

### Step 3: Substitute the expressions for \( a, b, c \)
Substituting the expressions from Step 1 into the inequalities:

1. **First Inequality**:
\[ a + b > c \implies a_1 + a_1 r > a_1 r^2 \]
Dividing through by \( a_1 \) (assuming \( a_1 > 0 \)):
\[ 1 + r > r^2 \implies r^2 - r - 1 < 0 \]

2. **Second Inequality**:
\[ a + c > b \implies a_1 + a_1 r^2 > a_1 r \]
Dividing through by \( a_1 \):
\[ 1 + r^2 > r \implies r^2 - r + 1 > 0 \]

3. **Third Inequality**:
\[ b + c > a \implies a_1 r + a_1 r^2 > a_1 \]
Dividing through by \( a_1 \):
\[ r + r^2 > 1 \implies r^2 + r - 1 > 0 \]

### Step 4: Solve the inequalities
1. **For the first inequality** \( r^2 - r - 1 < 0 \):
- Find the roots using the quadratic formula:
\[ r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
- The roots are \( r_1 = \frac{1 - \sqrt{5}}{2} \) and \( r_2 = \frac{1 + \sqrt{5}}{2} \).
- The inequality \( r^2 - r - 1 < 0 \) holds between the roots:
\[ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]

2. **For the second inequality** \( r^2 - r + 1 > 0 \):
- The discriminant \( D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0 \).
- This quadratic is always positive.

3. **For the third inequality** \( r^2 + r - 1 > 0 \):
- The roots are the same as the first inequality, so we have:
\[ r < \frac{1 - \sqrt{5}}{2} \quad \text{or} \quad r > \frac{1 + \sqrt{5}}{2} \]

### Step 5: Combine the results
The valid range for \( r \) must satisfy both conditions. Thus, we take the intersection of the ranges:
- From the first inequality: \( \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \)
- From the third inequality: \( r < \frac{1 - \sqrt{5}}{2} \) or \( r > \frac{1 + \sqrt{5}}{2} \)

The only valid range for \( r \) is:
\[ r \in \left( \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \right) \]

### Conclusion
The common ratio \( r \) of the sides of triangle ABC in G.P. must lie within the interval:
\[ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]