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 a geometric progression (G.P.) be \( a, ar, ar^2 \). By the triangle inequality, the sum of any two side lengths must exceed the third side. This yields:

• \( a + ar > ar^2 \), which simplifies to \( 1 + r > r^2 \) after dividing by \( a \).
• \( a + ar^2 > ar \), which simplifies to \( 1 + r^2 > r \) after dividing by \( a \).
• \( ar + ar^2 > a \), which simplifies to \( r + r^2 > 1 \) after dividing by \( a \).

Analyzing the inequality \( 1 + r > r^2 \), we rearrange it to \( r^2 - r - 1 < 0 \). The roots of \( r^2 - r - 1 = 0 \) are \( r = \frac{1 \pm \sqrt{5}}{2} \). Given \( r > 1 \), the inequality holds for \( 1 < r < \frac{1+\sqrt{5}}{2} \), which is approximately \( 1 < r < 1.618 \). The inequality \( 1 + r^2 > r \) is always true for \( r > 1 \) because \( r^2 \) grows faster than \( r \). Similarly, \( r + r^2 > 1 \) is also trivially satisfied for \( r > 1 \). Thus, the only significant inequality is \( r^2 - r - 1 < 0 \), which confirms:

• \( 1 < r < 1.618 \).

Next, we determine \( 3\lfloor r \rfloor + \lfloor -r \rfloor \). Since \( 1 < r < 1.618 \), we have \( \lfloor r \rfloor = 1 \). For \( -r \), its value lies between approximately \(-1.618\) and \(-1\), hence \( \lfloor -r \rfloor = -2 \).

Therefore, the calculation is:

• \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 3(1) + (-2) = 3 - 2 = 1 \).

Consequently, \( 3\lfloor r \rfloor + \lfloor -r \rfloor = 1 \). This result is consistent with the specified range of [1,1].