Sequences & Series
Sum of infinite GP
Grade 11
Question:
<p>If the sum of the series \( 1 - \dfrac{3}{x} - \dfrac{9}{x^2} - \dfrac{27}{x^3} - \ldots \) to \( \infty \) is a finite number, then</p>
<p>\( x < 3 \)</p>
<p>\( x > \dfrac{1}{3} \)</p>
<p>\( x < \dfrac{1}{3} \)</p>
<p>\( x > 3 \)</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric series with first term a = 1 and common ratio r = -3/x. For convergence, we need |r| < 1, which gives |x| > 3. The sum formula S = a/(1-r) applies only when this condition is satisfied.
<p><strong>Step 1:</strong> Identify the series structure</p><p>The series is: 1 - 3/x - 9/x² - 27/x³ - ... = 1 + (-3/x) + (-3/x)² + (-3/x)³ + ...</p><p><strong>Step 2:</strong> Recognize as geometric series</p><p>This is a geometric series with first term a = 1 and common ratio r = -3/x</p><p><strong>Step 3:</strong> Apply convergence condition</p><p>For the series to converge to a finite sum, we need |r| < 1</p><p>|−3/x| < 1</p><p>3/|x| < 1</p><p>3 < |x|</p><p>|x| > 3</p><p><strong>Step 4:</strong> Calculate the sum (when convergent)</p><p>S = a/(1 - r) = 1/(1 - (-3/x)) = 1/(1 + 3/x) = x/(x + 3)</p><p>∴ The series converges to a finite number when |x| > 3, and the sum equals x/(x + 3)</p>
Correct Answer: D