Question

If three successive terms of a G.P. with common ratio $r \, (r>1)$ are the lengths of the sides of a triangle and $\lfloor r \rfloor$ denotes the greatest integer less than or equal to $r$, then $3\lfloor r \rfloor + \lfloor -r \rfloor$ is equal to:

Answer 1

Correct Answer: 1

Let the three successive terms of the G.P. be \( a, ar, ar^2 \), where \( r \(>\) 1 \). Since these terms form the sides of a triangle, they must satisfy the triangle inequality:
\[ a + ar \(>\) ar^2, \quad ar + ar^2 \(>\) a, \quad a + ar^2 \(>\) ar \]

Checking the Triangle Inequality Conditions
1. \( a + ar \(>\) ar^2 \):
\[ a(1 + r) \(>\) ar^2 \implies 1 + r \(>\) r^2 \implies r^2 - r - 1 \(<\) 0 \]
Solving the quadratic inequality:
\[ r = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]
Since \( r \(>\) 1 \), we have:
\[ 1 \(<\) r \(<\) \frac{1 + \sqrt{5}}{2} \approx 1.618 \]

2. \( ar + ar^2 \(>\) a \):
\[ ar(1 + r) \(>\) a \implies r(1 + r) \(>\) 1 \]
This condition is always satisfied for \( r \(>\) 1 \).

3. \( a + ar^2 \(>\) ar \):
\[ a(1 + r^2) \(>\) ar \implies 1 + r^2 \(>\) r \implies r^2 - r + 1 \(>\) 0 \]
This condition is always true for all values of \( r \).

Finding the Value of \( \lfloor r \rfloor \)
Since \( 1 \(<\) r \(<\) \frac{1 + \sqrt{5}}{2} \approx 1.618 \), the greatest integer less than or equal to \( r \) is:
\[ \lfloor r \rfloor = 1 \]

Calculating \( 3\lfloor r \rfloor + \lfloor -r \rfloor \)
\[ 3\lfloor r \rfloor + \lfloor -r \rfloor = 3 \times 1 + \lfloor -r \rfloor \]
Since \( \lfloor -r \rfloor \) is the greatest integer less than or equal to \( -r \), and \( -1.618 \(<\) -r \(<\) -1 \), we have:
\[ \lfloor -r \rfloor = -2 \]
Thus:
\[ 3\lfloor r \rfloor + \lfloor -r \rfloor = 3 \times 1 + (-2) = 1 \]

Conclusion: \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 1 \).