<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 a₁r = 1, express sum S = a/(1-r) in terms of r alone, then minimize using calculus. The constraint |r| < 1 and S > 0 determines the feasible domain.
<p><strong>Step 1:</strong> Let first term = a and common ratio = r. Given: ar = 1, so a = 1/r.</p><p><strong>Step 2:</strong> For convergence, |r| < 1. Sum S = a/(1-r) = (1/r)/(1-r) = 1/[r(1-r)].</p><p><strong>Step 3:</strong> For S > 0, we need r(1-r) > 0, which means 0 < r < 1 OR r < 0 and r > 1 (impossible). But if r < 0, then 1-r > 1 > 0, so r(1-r) < 0 if r < 0. Thus we need 0 < r < 1 OR r ∈ (-1, 0).</p><p><strong>Step 4:</strong> When r ∈ (-1, 0): r(1-r) is negative, making S = 1/[r(1-r)] negative. So this case doesn't satisfy S > 0.</p><p><strong>Step 5:</strong> For r ∈ (0,1): S = 1/[r(1-r)]. To minimize, let f(r) = r(1-r). Maximum of f(r) gives minimum of S.</p><p><strong>Step 6:</strong> df/dr = 1 - 2r = 0 ⟹ r = 1/2. At r = 1/2: f(1/2) = (1/2)(1/2) = 1/4.</p><p><strong>Step 7:</strong> S_min = 1/(1/4) = 4.</p><p>∴ Answer: B</p>
Correct Answer: B