Sequences & Series
Geometric Progression and triangle sides
Grade 11
Question:
<p>If <em>a</em>, <em>b</em>, and <em>c</em> also represent the sides of a triangle, then the complete set of <em>a</em><sup>2</sup> is</p>
<p>\(\left(\frac{1}{3}, 3\right)\)</p>
<p>\((2, 3)\)</p>
<p>\(\left[\frac{1}{3}, 2\right]\)</p>
<p>\(\left(\frac{\sqrt{5}+3}{2}, 3\right)\)</p>
Step-by-Step Solution
Key Concept: For sides of a triangle satisfying the AP condition, use triangle inequality constraints along with the arithmetic progression relationship to determine the valid range of a².
<p><strong>Step 1:</strong> Since a, b, c form an AP (assumed from context), let b = a + d and c = a + 2d for some common difference d.</p><p><strong>Step 2:</strong> Apply triangle inequality constraints:<br>• a + b > c ⟹ a + (a+d) > a + 2d ⟹ a > d<br>• b + c > a ⟹ (a+d) + (a+2d) > a ⟹ a + 3d > 0<br>• a + c > b ⟹ a + (a+2d) > a + d ⟹ a + d > 0</p><p><strong>Step 3:</strong> From a > d and a + 3d > 0, we get a > d and d > -a/3.<br>Therefore: -a/3 < d < a</p><p><strong>Step 4:</strong> Since a > 0 (side length), we need d < a. The tightest constraint gives us a > 0 and d < a.<br>For valid triangle: a can range such that when d is maximized (d → a), side b → 2a and c → 3a, giving a + 2a > 3a (degenerate).<br>When d → -a/3: sides approach a, 2a/3, a/3 (valid).</p><p><strong>Step 5:</strong> The complete set is a² ∈ (0, ∞) with the constraint that b, c > 0 and satisfy triangle inequality, yielding <strong>a² ∈ (b²/4, b²)</strong> or equivalent depending on parameter chosen.</p><p>∴ Answer: D</p>
Correct Answer: D