Sequences & Series
HP and AP relations
Grade 11
Question:
<p>Given that <i>AD</i>, <i>BE</i>, <i>CF</i> are the altitudes of triangle <i>ABC</i> and are in HP, then which of the following is true?</p>
<p>\(a, b, c\) are in AP</p>
<p>\(\sin A, \sin B, \sin C\) are in AP</p>
<p>\(\sin A, \sin B, \sin C\) are in HP</p>
<p>\(\sin A, \sin B, \sin C\) are in GP</p>
Step-by-Step Solution
Key Concept: If altitudes are in HP, their reciprocals (proportional to sides via area formula) are in AP. Use the relationship: altitude = 2×Area/base to connect altitudes to sides.
<p><strong>Step 1:</strong> Let the altitudes be AD = h_a, BE = h_b, CF = h_c in HP.</p><p><strong>Step 2:</strong> If h_a, h_b, h_c are in HP, then 1/h_a, 1/h_b, 1/h_c are in AP.</p><p><strong>Step 3:</strong> Using the area formula: Area = (1/2) × base × altitude</p><p>We have: h_a = 2Δ/a, h_b = 2Δ/b, h_c = 2Δ/c (where Δ is area, a, b, c are opposite sides)</p><p><strong>Step 4:</strong> Therefore: 1/h_a = a/(2Δ), 1/h_b = b/(2Δ), 1/h_c = c/(2Δ)</p><p><strong>Step 5:</strong> Since 1/h_a, 1/h_b, 1/h_c are in AP: a, b, c are in AP</p><p><strong>Step 6:</strong> This means: 2b = a + c (the sides of the triangle are in AP)</p><p>∴ Answer: <strong>B</strong> (The sides a, b, c are in Arithmetic Progression)</p>
Correct Answer: B