<p>Let \(a_1, a_2, a_3, \ldots\) be in harmonic progression with \(a_1 = 5\) and \(a_{20} = 25\). The least positive integer \(n\) for which \(a_n < 0\) is</p>
Step-by-Step Solution
Key Concept: If a_1, a_2, a_3,... are in harmonic progression (HP), then 1/a_1, 1/a_2, 1/a_3,... form an arithmetic progression (AP). Use this to find the general term and determine when a_n < 0.
<p><strong>Step 1: Convert HP to AP</strong></p><p>If {a_n} is in HP, then {1/a_n} is in AP.</p><p>Let b_n = 1/a_n. Then {b_n} is an AP with:</p><p>b_1 = 1/a_1 = 1/5</p><p>b_20 = 1/a_20 = 1/25</p><p><strong>Step 2: Find the common difference of the AP</strong></p><p>For an AP: b_n = b_1 + (n-1)d</p><p>b_20 = b_1 + 19d</p><p>1/25 = 1/5 + 19d</p><p>19d = 1/25 - 1/5 = 1/25 - 5/25 = -4/25</p><p>d = -4/(25 × 19) = -4/475</p><p><strong>Step 3: Find the general term b_n</strong></p><p>b_n = 1/5 + (n-1)(-4/475)</p><p>b_n = 1/5 - 4(n-1)/475</p><p>b_n = (95 - 4(n-1))/475 = (95 - 4n + 4)/475 = (99 - 4n)/475</p><p><strong>Step 4: Find when a_n < 0</strong></p><p>Since a_n = 1/b_n, we need a_n < 0 when b_n < 0 (since we want the reciprocal to be negative).</p><p>b_n < 0 when: 99 - 4n < 0</p><p>99 < 4n</p><p>n > 99/4 = 24.75</p><p><strong>Step 5: Find the least positive integer n</strong></p><p>The least integer n satisfying n > 24.75 is n = 25.</p><p>Verification: b_25 = (99 - 100)/475 = -1/475 < 0, so a_25 < 0 ✓</p><p>b_24 = (99 - 96)/475 = 3/475 > 0, so a_24 > 0 ✓</p><p><strong>∴ Answer: D</strong></p>
Correct Answer: D