<p>Let \(a_n\) be an infinite geometric sequence with a convergent and negative sum. The common ratio of the sequence is \(r\) and the first term is \(a_1\), then which one of the following is always true?</p>
Step-by-Step Solution
Key Concept: For a geometric series to converge with negative sum, we need |r| < 1 for convergence AND the sum formula S = a₁/(1-r) must be negative, which constrains both the sign of a₁ and the value of r relative to 1.
<p><strong>Step 1:</strong> For convergence of a geometric series, we require |r| < 1, so -1 < r < 1.</p><p><strong>Step 2:</strong> The sum of a convergent geometric series is S = a₁/(1-r). Since 0 < 1-r < 2 when -1 < r < 1, the denominator (1-r) is always positive.</p><p><strong>Step 3:</strong> For S < 0 with (1-r) > 0, we must have a₁ < 0.</p><p><strong>Step 4:</strong> Verify: If a₁ < 0 and -1 < r < 1, then S = a₁/(1-r) is always negative (negative numerator, positive denominator).</p><p><strong>Step 5:</strong> Check the converse: If a₁ > 0, even with -1 < r < 1, we get S > 0, contradicting the given condition.</p><p>∴ Answer: The statement that is always true must involve <strong>a₁ < 0</strong> (without additional constraints on r beyond |r| < 1)</p>
Correct Answer: C