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

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: $3\lfloor r \rfloor + \lfloor -r \rfloor = 1$.