<p>Find the sum of all 3-digit natural numbers which are of the form \(3m + 2\), \(m \in \mathbb{N}\), i.e., leaves the remainder 2 when divided by 3.</p>
Step-by-Step Solution
Key Concept: 3-digit numbers leaving remainder 2 when divided by 3 form an arithmetic sequence with first term 101 and common difference 3. Use the AP sum formula S_n = n/2(first + last) after identifying the number of terms.
<p><strong>Step 1:</strong> Identify the sequence. Numbers of form 3m + 2 are: 2, 5, 8, 11, ..., leaving remainder 2 when divided by 3. For 3-digit numbers, we need 100 ≤ 3m + 2 ≤ 999.</p><p><strong>Step 2:</strong> Find first 3-digit term. From 3m + 2 ≥ 100: m ≥ 32.67, so m = 33, giving first term a = 3(33) + 2 = <strong>101</strong>.</p><p><strong>Step 3:</strong> Find last 3-digit term. From 3m + 2 ≤ 999: m ≤ 332.33, so m = 332, giving last term l = 3(332) + 2 = <strong>998</strong>.</p><p><strong>Step 4:</strong> This is an AP with first term a = 101, last term l = 998, common difference d = 3. Number of terms: n = (998 - 101)/3 + 1 = 897/3 + 1 = 299 + 1 = <strong>300</strong>.</p><p><strong>Step 5:</strong> Apply sum formula: S_n = n/2(a + l) = 300/2(101 + 998) = 150 × 1099 = <strong>164,850</strong>.</p><p>∴ Answer: <strong>164,850</strong></p>
Correct Answer: 164