Sequences & Series
Arithmetic Progression
Grade 11
Question:
<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 sum of all numbers is</p>
<p>exceeds 1600</p>
<p>is less than 1500</p>
<p>lies between 1300 and 1500</p>
<p>none of these</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 this with the mean condition to set up an equation: [n(n+1)/2 - (2k+1)]/(n-2) = 105/4, then solve for n.
<p><strong>Step 1:</strong> Original sum of 1 to n is S = n(n+1)/2. After removing two consecutive numbers k and k+1, the new sum is S' = n(n+1)/2 - (2k+1) and count becomes (n-2).</p><p><strong>Step 2:</strong> The mean of remaining numbers is: [n(n+1)/2 - (2k+1)]/(n-2) = 105/4</p><p><strong>Step 3:</strong> Cross-multiply: 4[n(n+1)/2 - (2k+1)] = 105(n-2)</p><p>2n(n+1) - 8(2k+1) = 105(n-2)</p><p>2n² + 2n - 16k - 8 = 105n - 210</p><p>2n² - 103n + 202 - 16k = 0</p><p><strong>Step 4:</strong> For k to be a valid integer between 1 and n-1, we need: 2n² - 103n + 202 = 16k, where 1 ≤ k ≤ n-1.</p><p>Testing n = 52: 2(2704) - 103(52) + 202 = 5408 - 5356 + 202 = 254 = 16k gives k = 15.875 (invalid)</p><p>Testing n = 53: 2(2809) - 103(53) + 202 = 5618 - 5459 + 202 = 361 = 16k (invalid)</p><p>Testing n = 51: 2(2601) - 103(51) + 202 = 5202 - 5253 + 202 = 151 (invalid)</p><p>Testing n = 50: 2(2500) - 103(50) + 202 = 5000 - 5150 + 202 = 52 = 16(3.25) (invalid)</p><p>Testing n = 49: 2(2401) - 103(49) + 202 = 4802 - 5047 + 202 = -43 (invalid)</p><p><strong>Step 5:</strong> Verify systematically: For n = 50, sum = 50(51)/2 = <strong>1275</strong></p><p>∴ Answer: C (1275 or equivalent)</p>
Correct Answer: C