<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 removed numbers</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). Setting up the equation for the new mean as 105/4 gives us two unknowns to solve simultaneously.
<p><strong>Step 1:</strong> Original sum of 1, 2, 3, ..., n is S = n(n+1)/2</p><p><strong>Step 2:</strong> Let the two consecutive removed numbers be k and k+1. Sum of removed numbers = 2k+1</p><p><strong>Step 3:</strong> Sum of remaining (n-2) numbers = n(n+1)/2 - (2k+1)</p><p><strong>Step 4:</strong> Given that arithmetic mean of remaining numbers is 105/4:</p><p>[n(n+1)/2 - (2k+1)]/(n-2) = 105/4</p><p><strong>Step 5:</strong> Simplifying: 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 k must be a positive integer with 1 ≤ k < n, test values. For n = 52: 2(2704) - 103(52) + 206 = 5408 - 5356 + 206 = 258 = 8k, so k = 32.25 (not integer)</p><p>For n = 51: 2(2601) - 103(51) + 206 = 5202 - 5253 + 206 = 155 (not divisible by 8)</p><p>For n = 50: 2(2500) - 103(50) + 206 = 5000 - 5150 + 206 = 56 = 8k, so k = 7 ✓</p><p><strong>Step 7:</strong> Verification: Sum = 50(51)/2 = 1275; Remove 7,8 (sum=15); Remaining sum = 1260; Mean = 1260/48 = 105/4 ✓</p><p>∴ The removed numbers are <strong>7 and 8</strong></p>
Correct Answer: A