<p>The total number of ways in which three distinct numbers in A.P. can be selected from the set \(\{1, 2, 3, \ldots, 24\}\) is equal to</p>
Step-by-Step Solution
Key Concept: For three numbers in A.P., if the first term is 'a' and common difference is 'd', they are a, a+d, a+2d. The constraint is that all three must lie in {1,2,...,24}, so a+2d ≤ 24. Count valid pairs (a,d) where d ≥ 1 and a ≥ 1.
<p><strong>Step 1:</strong> Represent three numbers in A.P. as a, a+d, a+2d where a is the first term and d is the common difference (d ≥ 1 for distinct numbers).</p><p><strong>Step 2:</strong> Apply constraints: All three numbers must be in {1,2,...,24}, so:</p><ul><li>a ≥ 1</li><li>a+2d ≤ 24</li></ul><p><strong>Step 3:</strong> For each value of d, count valid values of a:</p><ul><li>d=1: a can be 1 to 22 → 22 ways</li><li>d=2: a can be 1 to 20 → 20 ways</li><li>d=3: a can be 1 to 18 → 18 ways</li><li>...</li><li>d=11: a can be 1 to 2 → 2 ways</li></ul><p><strong>Step 4:</strong> For d=k, we need a+2k ≤ 24, so a ≤ 24-2k. Thus a ∈ {1,2,...,24-2k}, giving (24-2k) valid values.</p><p><strong>Step 5:</strong> Sum over all valid d: d ranges from 1 to 11 (since when d=12, we get a ≤ 0).</p><p>Total = 22+20+18+16+14+12+10+8+6+4+2 = 2(11+10+9+8+7+6+5+4+3+2+1) = 2 × 66 = 132</p><p>Alternatively: ∑(24-2k) for k=1 to 11 = 2∑(12-k) for k=1 to 11 = 2(11+10+...+1) = 2 × (11×12)/2 = 132</p><p>∴ Answer: <strong>132</strong></p>
Correct Answer: C