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, we need to analyze the sides of triangle ABC, denoted as \( a, b, c \), which are in geometric progression (G.P.) with a common ratio \( r \).

### Step-by-Step Solution:

1. **Express the sides in terms of the common ratio**:
Since \( a, b, c \) are in G.P., we can express them as:
\[ a = x, \quad b = xr, \quad c = xr^2 \]
where \( x \) is a positive constant.

2. **Apply the triangle inequality**:
For any triangle, the triangle inequalities must hold:
- \( a + b > c \)
- \( a + c > b \)
- \( b + c > a \)

Substituting the expressions for \( a, b, c \):
- \( x + xr > xr^2 \)
- \( x + xr^2 > xr \)
- \( xr + xr^2 > x \)

3. **Simplify the inequalities**:
- From \( x + xr > xr^2 \):
\[ x(1 + r) > xr^2 \implies 1 + r > r^2 \implies r^2 - r - 1 < 0 \]
- From \( x + xr^2 > xr \):
\[ x(1 + r^2) > xr \implies 1 + r^2 > r \implies r^2 - r + 1 > 0 \quad \text{(always true)} \]
- From \( xr + xr^2 > x \):
\[ x(r + r^2) > x \implies r + r^2 > 1 \implies r^2 + r - 1 > 0 \]

4. **Solve the quadratic inequalities**:
- For \( r^2 - r - 1 < 0 \):
The roots can be found using the quadratic formula:
\[ r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]
This gives us the interval:
\[ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]

- For \( r^2 + r - 1 > 0 \):
The roots are:
\[ r = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]
This gives us the intervals:
\[ r < \frac{-1 - \sqrt{5}}{2} \quad \text{or} \quad r > \frac{-1 + \sqrt{5}}{2} \]

5. **Find the intersection of the intervals**:
The valid range for \( r \) must satisfy both inequalities:
- From \( r^2 - r - 1 < 0 \):
\[ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]
- From \( r^2 + r - 1 > 0 \):
\[ r > \frac{-1 + \sqrt{5}}{2} \]

Therefore, the intersection of these intervals is:
\[ \frac{-1 + \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]

### Conclusion:
The common ratio \( r \) must lie within the interval:
\[ \left(\frac{-1 + \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\right) \]