<p><strong>41.</strong> The sum of 20 terms of a series of which every even term is 2 times the term before it, and every odd term is 3 times the term before it, the first term being unity is</p>
<p>\(\left(\dfrac{2}{7}\right)(6^{10}-1)\)</p>
<p>\(\left(\dfrac{3}{7}\right)(6^{10}-1)\)</p>
<p>\(\left(\dfrac{3}{5}\right)(6^{10}-1)\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Identify the pattern by separating odd and even positioned terms: odd positions follow one recurrence (multiply by 3), even positions follow another (multiply by 2). Sum each subsequence separately using geometric series formula.
<p><strong>Step 1:</strong> Write out the pattern with first term = 1:<br>T₁ = 1, T₂ = 2·1 = 2, T₃ = 3·2 = 6, T₄ = 2·6 = 12, T₅ = 3·12 = 36, ...</p><p><strong>Step 2:</strong> Separate into odd and even positioned terms:<br>Odd positions (1st, 3rd, 5th, ...): 1, 6, 36, 216, ... → each term is 6 times previous<br>Even positions (2nd, 4th, 6th, ...): 2, 12, 72, ... → each term is 6 times previous</p><p><strong>Step 3:</strong> Identify the sequences:<br>Odd terms: T₁ = 1, T₃ = 1·6, T₅ = 1·6², ..., T₁₉ = 1·6⁹ (10 terms with first term 1, ratio 6)<br>Even terms: T₂ = 2, T₄ = 2·6, T₆ = 2·6², ..., T₂₀ = 2·6⁹ (10 terms with first term 2, ratio 6)</p><p><strong>Step 4:</strong> Apply geometric series formula S = a(rⁿ - 1)/(r - 1):<br>Sum of odd terms = 1·(6¹⁰ - 1)/(6 - 1) = (6¹⁰ - 1)/5<br>Sum of even terms = 2·(6¹⁰ - 1)/(6 - 1) = 2(6¹⁰ - 1)/5</p><p><strong>Step 5:</strong> Total sum = (6¹⁰ - 1)/5 + 2(6¹⁰ - 1)/5 = 3(6¹⁰ - 1)/5</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B