<p>The sum of the first 20 terms common between the series \(3 + 7 + 11 + 15 + \cdots\) and \(1 + 6 + 11 + 16 + \cdots\), is</p>
Step-by-Step Solution
Key Concept: Find the common terms by determining which elements satisfy both arithmetic progressions simultaneously, then recognize that common terms themselves form an AP with a specific common difference equal to the LCM of the original common differences.
<p><strong>Step 1:</strong> Write both series in standard form.</p><p>Series 1: $a_n = 3 + (n-1)·4 = 4n - 1$</p><p>Series 2: $b_m = 1 + (m-1)·5 = 5m - 4$</p><p><strong>Step 2:</strong> Find common terms by setting $4n - 1 = 5m - 4$, which gives $4n = 5m - 3$.</p><p>This means $4n \equiv -3 \pmod{5}$, so $n \equiv 3 \pmod{5}$.</p><p>Thus $n = 5k + 3$ for non-negative integers $k$.</p><p><strong>Step 3:</strong> Common terms form an AP: substituting $n = 5k + 3$ into Series 1:</p><p>$c_k = 4(5k + 3) - 1 = 20k + 11$</p><p>Common difference = 20, first term = 11.</p><p><strong>Step 4:</strong> Sum of first 20 common terms:</p><p>$S_{20} = \frac{20}{2}[2(11) + 19(20)] = 10[22 + 380] = 10(402) = 4020$</p><p>∴ Answer: D</p>
Correct Answer: D