<p><strong>40.</strong> Let \(a_n\) be the \(n\)th term of a G.P. of positive numbers. Let \(\displaystyle\sum_{n=1}^{100} a_{2n} = \alpha\) and \(\displaystyle\sum_{n=1}^{100} a_{2n-1} = \beta\), such that \(\alpha \neq \beta\), then the common ratio is</p>
Step-by-Step Solution
Key Concept: In a G.P., separate the series into odd-indexed and even-indexed terms, then use their sum formulas to find the common ratio by taking their ratio.
<p><strong>Step 1:</strong> Let the G.P. have first term <em>a</em> and common ratio <em>r</em> (r > 0).</p><p><strong>Step 2:</strong> Odd-indexed terms: a₁, a₃, a₅, ... = a, ar², ar⁴, ... form a G.P. with first term <em>a</em> and common ratio <em>r</em>²</p><p>β = Σₙ₌₁¹⁰⁰ a₂ₙ₋₁ = a(1 - r²⁰⁰)/(1 - r²)</p><p><strong>Step 3:</strong> Even-indexed terms: a₂, a₄, a₆, ... = ar, ar³, ar⁵, ... form a G.P. with first term <em>ar</em> and common ratio <em>r</em>²</p><p>α = Σₙ₌₁¹⁰⁰ a₂ₙ = ar(1 - r²⁰⁰)/(1 - r²)</p><p><strong>Step 4:</strong> Take the ratio: α/β = [ar(1 - r²⁰⁰)/(1 - r²)] / [a(1 - r²⁰⁰)/(1 - r²)] = ar/a = <em>r</em></p><p>∴ Answer: A</p>
Correct Answer: A