<p>If three successive terms of a G.P. having common ratio <i>r</i> form the sides of a triangle, then the possible value(s) of \([r]\) is/are (where \([\cdot]\) denotes greatest integer function)</p>
Step-by-Step Solution
Key Concept: Apply triangle inequality to three terms in G.P.: the sum of any two must exceed the third.
<p><strong>Step 1:</strong> Let three successive terms of G.P. be \(\frac{a}{r}, a, ar\).<br/><strong>Step 2:</strong> For triangle inequality, sum of any two sides > third side.<br/>\(\frac{a}{r} + a > ar \Rightarrow \frac{1}{r} + 1 > r \Rightarrow 1 + r > r^2\)<br/>\(a + ar > \frac{a}{r} \Rightarrow 1 + r > \frac{1}{r} \Rightarrow r(1+r) > 1 \Rightarrow r^2 + r - 1 > 0\)<br/>\(\frac{a}{r} + ar > a \Rightarrow \frac{1}{r} + r > 1\)<br/><strong>Step 3:</strong> From \(r^2 + r - 1 > 0\): \(r > \frac{-1+\sqrt{5}}{2} \approx 0.618\)<br/>From \(r^2 - r - 1 < 0\): \(r < \frac{1+\sqrt{5}}{2} \approx 1.618\)<br/>Thus \(0.618 < r < 1.618\), so \([r] = 0\) or \([r] = 1\).<br/>For \(r > 0\), we have \([r] \in \{0, 1\}\). The answer includes 0.</p>
Correct Answer: B