Three successive terms of a G.P. will form the sides of a triangle if the common ratio r satisfies the inequality

Options

A. `(sqrt(3)-1)/(2)ltrlt(sqrt(3)+1)/(2)`
B. `(sqrt(5)-1)/(2)ltrlt(sqrt(5)+1)/(2)`
C. `(sqrt(2)-1)/(2)ltrlt(sqrt(2)+1)/(2)`
D. none of these

Explanation

To determine the conditions under which three successive terms of a geometric progression (G.P.) can form the sides of a triangle, we can denote the terms as \( a \), \( ar \), and \( ar^2 \), where \( a \) is a positive constant and \( r \) is the common ratio.

### Step 1: Identify the terms of the G.P.
Let the three terms of the G.P. be:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

### Step 2: Apply the triangle inequality
For these terms to form the sides of a triangle, they must satisfy the triangle inequality:
1. \( a + ar > ar^2 \)
2. \( a + ar^2 > ar \)
3. \( ar + ar^2 > a \)

### Step 3: Simplify the inequalities
Let's simplify each inequality:

**Inequality 1:**
\[ a + ar > ar^2 \implies a(1 + r) > ar^2 \implies 1 + r > r^2 \implies r^2 - r - 1 < 0 \]

**Inequality 2:**
\[ a + ar^2 > ar \implies a(1 + r^2) > ar \implies 1 + r^2 > r \implies r^2 - r + 1 > 0 \]
(This inequality is always true for all real \( r \) since the discriminant is negative.)

**Inequality 3:**
\[ ar + ar^2 > a \implies a(r + r^2) > a \implies r + r^2 > 1 \implies r^2 + r - 1 > 0 \]

### Step 4: Solve the quadratic inequalities
Now, we need to solve the inequalities \( r^2 - r - 1 < 0 \) and \( r^2 + r - 1 > 0 \).

**For \( r^2 - r - 1 < 0 \):**
The roots of the equation \( r^2 - r - 1 = 0 \) can be found using the quadratic formula:
\[ r = \frac{1 \pm \sqrt{5}}{2} \]
This gives us the roots \( r_1 = \frac{1 - \sqrt{5}}{2} \) and \( r_2 = \frac{1 + \sqrt{5}}{2} \).

The quadratic \( r^2 - r - 1 < 0 \) is negative between its roots:
\[ \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]

**For \( r^2 + r - 1 > 0 \):**
The roots of the equation \( r^2 + r - 1 = 0 \) can also be found using the quadratic formula:
\[ r = \frac{-1 \pm \sqrt{5}}{2} \]
This gives us the roots \( r_3 = \frac{-1 - \sqrt{5}}{2} \) and \( r_4 = \frac{-1 + \sqrt{5}}{2} \).

The quadratic \( r^2 + r - 1 > 0 \) is positive outside its roots:
\[ r < \frac{-1 - \sqrt{5}}{2} \quad \text{or} \quad r > \frac{-1 + \sqrt{5}}{2} \]

### Step 5: Combine the intervals
Now, we need to combine the intervals from both inequalities:
1. From \( r^2 - r - 1 < 0 \): \( \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \)
2. From \( r^2 + r - 1 > 0 \): \( r < \frac{-1 - \sqrt{5}}{2} \) or \( r > \frac{-1 + \sqrt{5}}{2} \)

Since \( \frac{-1 - \sqrt{5}}{2} \) is negative and irrelevant for our case (as \( r \) must be positive), we only consider \( r > \frac{-1 + \sqrt{5}}{2} \).

### Final Interval
Thus, the valid interval for \( r \) that satisfies both conditions is:
\[ \frac{-1 + \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \]