<p>Find the sum of the first hundred numbers appearing in both the progressions 17, 21, 25, … and 16, 21, 26, …</p>
Step-by-Step Solution
Key Concept: Find the common terms by identifying the general forms of both progressions, then use the Chinese Remainder Theorem concept to find terms satisfying both conditions simultaneously. The common terms themselves form an arithmetic progression.
<p><strong>Step 1:</strong> Write general forms of both progressions.</p><p>First progression: <strong>17, 21, 25, ...</strong> → General term: <strong>aₙ = 17 + 4(n-1) = 4n + 13</strong></p><p>Second progression: <strong>16, 21, 26, ...</strong> → General term: <strong>bₘ = 16 + 5(m-1) = 5m + 11</strong></p><p><strong>Step 2:</strong> Find when terms are equal: <strong>4n + 13 = 5m + 11</strong></p><p>This gives: <strong>4n = 5m - 2</strong> or <strong>4n ≡ -2 (mod 5)</strong></p><p>Since <strong>4n ≡ -n (mod 5)</strong>, we get <strong>-n ≡ -2 (mod 5)</strong>, so <strong>n ≡ 2 (mod 5)</strong></p><p><strong>Step 3:</strong> Common terms occur when <strong>n = 2, 7, 12, 17, ...</strong> (i.e., n = 5k + 2)</p><p>Substituting into first progression: <strong>4(5k + 2) + 13 = 20k + 21</strong></p><p>Common terms form AP: <strong>21, 41, 61, 81, ...</strong> with first term <strong>a = 21</strong> and common difference <strong>d = 20</strong></p><p><strong>Step 4:</strong> Sum of first 100 terms:</p><p><strong>S₁₀₀ = (100/2)[2(21) + 99(20)]</strong></p><p><strong>S₁₀₀ = 50[42 + 1980]</strong></p><p><strong>S₁₀₀ = 50 × 2022 = 101,100</strong></p><p>∴ <strong>Answer: 101,100</strong></p>
Correct Answer: 101