<p>Let \(S\) denote the sum of an infinite geometric sequence with \(S > 0\). If the second term of this sequence is 1, then the minimum possible value of \(S\) is:</p>
Step-by-Step Solution
Key Concept: For a convergent geometric series with second term ar = 1, express S in terms of first term a and use calculus to minimize S = a/(1-r) subject to the constraint that |r| < 1 and ar = 1.
<p><strong>Step 1:</strong> Set up the constraint. For geometric sequence with first term a and common ratio r: second term = ar = 1, so a = 1/r.</p><p><strong>Step 2:</strong> Write sum S = a/(1-r) = (1/r)/(1-r) = 1/(r(1-r)). For convergence, |r| < 1 and r ≠ 0.</p><p><strong>Step 3:</strong> Since S > 0, analyze the sign: r(1-r) > 0 requires 0 < r < 1 (if r < 0, then r(1-r) < 0, making S < 0).</p><p><strong>Step 4:</strong> Minimize S = 1/(r(1-r)) for r ∈ (0,1). Equivalently, maximize f(r) = r(1-r) = r - r².</p><p><strong>Step 5:</strong> Take derivative: f'(r) = 1 - 2r = 0, giving r = 1/2.</p><p><strong>Step 6:</strong> At r = 1/2: f(1/2) = (1/2)(1/2) = 1/4, so S_min = 1/(1/4) = 4.</p><p><strong>Verification:</strong> When r = 1/2, a = 2. Sequence: 2, 1, 1/2, 1/4, ... with sum = 2/(1-1/2) = 4.</p><p>∴ Answer: B (minimum value = 4)</p>
Correct Answer: B