<p>Find the sum of all numbers from 100 to 550 which are divisible by 9.</p>
Step-by-Step Solution
Key Concept: Identify divisible numbers as an arithmetic progression (AP) where first term is the smallest multiple of 9 ≥ 100 and last term is the largest multiple of 9 ≤ 550, then use AP sum formula with common difference 9.
<p><strong>Step 1:</strong> Find the first number ≥ 100 divisible by 9.</p><p>100 ÷ 9 = 11.11..., so first term = 9 × 12 = 108</p><p><strong>Step 2:</strong> Find the last number ≤ 550 divisible by 9.</p><p>550 ÷ 9 = 61.11..., so last term = 9 × 61 = 549</p><p><strong>Step 3:</strong> These form an AP: 108, 117, 126, ..., 549 with common difference d = 9</p><p><strong>Step 4:</strong> Find number of terms using aₙ = a + (n-1)d</p><p>549 = 108 + (n-1)9</p><p>441 = (n-1)9</p><p>n - 1 = 49, so n = 50</p><p><strong>Step 5:</strong> Apply sum formula: S = n/2(first term + last term)</p><p>S = 50/2(108 + 549) = 25 × 657 = 16,425</p><p>∴ Answer: <strong>16,425</strong></p>
Correct Answer: 16