<p><strong>For Problems 22–24:</strong> Two consecutive numbers from 1, 2, 3, …, \(n\) are removed. The arithmetic mean of the remaining numbers is \(\frac{105}{4}\).</p><p>The value of \(n\) lies in</p>
Step-by-Step Solution
Key Concept: When two consecutive numbers k and k+1 are removed from 1,2,...,n, the sum decreases by (2k+1). Use the mean formula for remaining (n-2) numbers to set up an equation involving k and n.
<p><strong>Step 1:</strong> The sum of numbers from 1 to n is <strong>S = n(n+1)/2</strong></p><p><strong>Step 2:</strong> Let the two consecutive numbers removed be k and k+1, where 1 ≤ k ≤ n-1. Sum removed = k + (k+1) = 2k+1</p><p><strong>Step 3:</strong> Sum of remaining numbers = n(n+1)/2 - (2k+1)</p><p><strong>Step 4:</strong> Mean of remaining (n-2) numbers is given as 105/4:</p><p>[n(n+1)/2 - (2k+1)]/(n-2) = 105/4</p><p><strong>Step 5:</strong> Simplify: n(n+1)/2 - (2k+1) = 105(n-2)/4</p><p>2n(n+1) - 4(2k+1) = 105(n-2)</p><p>2n² + 2n - 8k - 4 = 105n - 210</p><p>2n² - 103n + 206 = 8k</p><p><strong>Step 6:</strong> Since 1 ≤ k ≤ n-1, we need: 1 ≤ (2n² - 103n + 206)/8 ≤ n-1</p><p><strong>Step 7:</strong> From the lower bound: 2n² - 103n + 206 ≥ 8, so 2n² - 103n + 198 ≥ 0</p><p>Using the quadratic formula: n ≥ 51 or n ≤ 1.94 (rejected since n > 1)</p><p><strong>Step 8:</strong> From the upper bound: 2n² - 103n + 206 ≤ 8(n-1), so 2n² - 111n + 214 ≤ 0</p><p>Using the quadratic formula: 2 ≤ n ≤ 53.5</p><p><strong>Step 9:</strong> Combining: 51 ≤ n ≤ 53</p><p>∴ Answer: A</p>
Correct Answer: A