If three successive terms of G.P from the sides of a triangle then show that common ratio ' r' satisfies the inequality `1/2 (sqrt(5)-1) lt r lt 1/2 (sqrt(5)+1)` .

Answer

To show that the common ratio \( r \) of three successive terms of a geometric progression (G.P.) that form the sides of a triangle satisfies the inequality

\[ \frac{1}{2}(\sqrt{5}-1) < r < \frac{1}{2}(\sqrt{5}+1), \]

we will analyze two cases based on the value of \( r \).

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

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

### Step 3: Simplify the inequalities
#### Case 1: Assume \( r > 1 \)
In this case, \( ar^2 \) is the largest side. The first inequality becomes:
\[ a + ar > ar^2 \]
Factoring out \( a \) gives:
\[ a(1 + r) > ar^2 \]
Dividing both sides by \( a \) (assuming \( a > 0 \)):
\[ 1 + r > r^2 \]
Rearranging gives:
\[ r^2 - r - 1 < 0 \]
Now, we find the roots of the quadratic equation \( r^2 - r - 1 = 0 \) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ r_1 = \frac{1 - \sqrt{5}}{2}, \quad r_2 = \frac{1 + \sqrt{5}}{2} \]
Since \( r > 1 \), we only consider the interval:
\[ 1 < r < \frac{1 + \sqrt{5}}{2} \]

#### Case 2: Assume \( r < 1 \)
In this case, \( a \) is the largest side. The second inequality becomes:
\[ a + ar^2 > ar \]
Factoring out \( a \) gives:
\[ a(1 + r^2) > ar \]
Dividing both sides by \( a \) (assuming \( a > 0 \)):
\[ 1 + r^2 > r \]
Rearranging gives:
\[ r^2 - r + 1 > 0 \]
This inequality is always true for all real \( r \).

Now, we analyze the third inequality:
\[ ar + ar^2 > a \]
Factoring out \( a \) gives:
\[ a(r + r^2) > a \]
Dividing both sides by \( a \) (assuming \( a > 0 \)):
\[ r + r^2 > 1 \]
Rearranging gives:
\[ r^2 + r - 1 > 0 \]
Finding the roots of \( r^2 + r - 1 = 0 \):
\[ r = \frac{-1 \pm \sqrt{5}}{2} \]
Thus, the roots are:
\[ r_1 = \frac{-1 - \sqrt{5}}{2}, \quad r_2 = \frac{-1 + \sqrt{5}}{2} \]
Since we are considering \( r < 1 \), we have:
\[ \frac{-1 + \sqrt{5}}{2} < r < 1 \]

### Step 4: Combine the results
From both cases, we have:
1. For \( r > 1 \): \( 1 < r < \frac{1 + \sqrt{5}}{2} \)
2. For \( r < 1 \): \( \frac{-1 + \sqrt{5}}{2} < r < 1 \)

Thus, combining these intervals gives us:
\[ \frac{1}{2}(\sqrt{5}-1) < r < \frac{1}{2}(\sqrt{5}+1) \]

### Conclusion
We have shown that if three successive terms of a G.P. form the sides of a triangle, the common ratio \( r \) satisfies the inequality:
\[ \frac{1}{2}(\sqrt{5}-1) < r < \frac{1}{2}(\sqrt{5}+1). \]