<p>Number of ways in which three numbers in A.P. can be selected from 1, 2, 3, …, <i>n</i> is</p>
<p>\(\left(\dfrac{n-1}{2}\right)^2\) if <i>n</i> is even</p>
<p>\(\dfrac{n(n-2)}{4}\) if <i>n</i> is even</p>
<p>\(\dfrac{(n-1)^2}{4}\) if <i>n</i> is odd</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Three numbers in A.P. can be written as (a-d, a, a+d) where a is the middle term and d is the common difference. Count valid pairs (a,d) such that 1 ≤ a-d and a+d ≤ n.
<p><strong>Step 1:</strong> Express three numbers in A.P. as (a-d, a, a+d) where a is the middle term and d > 0 is the common difference.</p><p><strong>Step 2:</strong> For valid selection from {1, 2, ..., n}, we need:</p><ul><li>a - d ≥ 1, so a ≥ d + 1</li><li>a + d ≤ n, so a ≤ n - d</li><li>d ≥ 1 (for three distinct numbers)</li></ul><p><strong>Step 3:</strong> For each value of d, the middle term a can range from (d+1) to (n-d). This is valid when d + 1 ≤ n - d, giving d ≤ (n-1)/2.</p><p><strong>Step 4:</strong> For each d from 1 to ⌊(n-1)/2⌋, the number of valid values of a is: (n-d) - (d+1) + 1 = n - 2d.</p><p><strong>Step 5:</strong> Total count = Σ(n - 2d) for d = 1 to ⌊(n-1)/2⌋.</p><p><strong>Case 1 (n odd, n = 2m+1):</strong> Sum = Σ(2m+1-2d) for d=1 to m = 2m + (2m-2) + (2m-4) + ... + 2 = 2(m + (m-1) + ... + 1) = m(m+1) = ((n-1)/2)((n+1)/2) = (n²-1)/4</p><p><strong>Case 2 (n even, n = 2m):</strong> Sum = Σ(2m-2d) for d=1 to m-1 = 2(m-1) + 2(m-2) + ... + 2 = 2((m-1) + (m-2) + ... + 1) = (m-1)m = ((n/2)-1)(n/2) = (n²-4)/4</p><p>∴ Answer: <strong>BC</strong> (representing both formulas for odd and even n respectively)</p>
Correct Answer: BC