Sequences & Series
Harmonic Progression
Grade 11
Question:
<p>Let \(\dfrac{1}{x_1}, \dfrac{1}{x_2}, \ldots, \dfrac{1}{x_n}\) (\(x_i \neq 0\) for \(i = 1, 2, \ldots, n\)) be in AP such that \(x_1 = 4\) and \(x_{21} = 20\). If <i>n</i> is the least positive integer for which \(x_n > 50\), then \(\displaystyle\sum_{i=1}^{n}\left(\dfrac{1}{x_i}\right)\) is equal to</p>
<p>\(\dfrac{1}{8}\)</p>
<p>3</p>
<p>\(\dfrac{13}{8}\)</p>
<p>\(\dfrac{13}{4}\)</p>
Step-by-Step Solution
Key Concept: If 1/x₁, 1/x₂, ..., 1/xₙ are in AP with first term 1/4 and 21st term 1/20, use the AP formula to find the common difference, then determine the smallest n where xₙ > 50 (equivalently, 1/xₙ < 1/50), and finally sum the AP series.
<p><strong>Step 1:</strong> Since 1/x₁, 1/x₂, ..., 1/xₙ form an AP, let this AP have first term a = 1/x₁ = 1/4 and common difference d.</p><p><strong>Step 2:</strong> Use the 21st term: 1/x₂₁ = a + 20d = 1/4 + 20d = 1/20</p><p>Solving: 20d = 1/20 - 1/4 = 1/20 - 5/20 = -4/20 = -1/5</p><p>∴ d = -1/100</p><p><strong>Step 3:</strong> The general term is 1/xₙ = 1/4 + (n-1)(-1/100) = 1/4 - (n-1)/100 = (25 - n + 1)/100 = (26 - n)/100</p><p><strong>Step 4:</strong> Therefore xₙ = 100/(26 - n)</p><p>For xₙ > 50: 100/(26 - n) > 50</p><p>100 > 50(26 - n)</p><p>100 > 1300 - 50n</p><p>50n > 1200</p><p>n > 24</p><p>∴ n = 25 (least positive integer)</p><p><strong>Step 5:</strong> Sum of AP: Sₙ = n/2[2a + (n-1)d] = 25/2[2(1/4) + 24(-1/100)]</p><p>= 25/2[1/2 - 24/100] = 25/2[1/2 - 6/25] = 25/2[(25 - 12)/50] = 25/2 × 13/50 = 325/100 = 13/4</p><p>∴ Answer: C (13/4)</p>
Correct Answer: C