Sequences & Series
Geometric Progression
Grade 11
Question:
<p>Let the GP be \(a, ar, ar^2, \ldots (0 < r < 1)\). From the question, \(\dfrac{a}{1-r} = 3^3 + 3 \cdot 3 - 9\) and \(a - ar = 3\). Find \(p + q\) where \(p\) and \(q\) are determined from the solution \(r = \dfrac{2}{3}, \dfrac{4}{3}\) giving \(p = 2, q = 3\). What is \(p + q\)?</p>
Step-by-Step Solution
Key Concept: For a GP with first term a and common ratio r where 0 < r < 1, the sum to infinity converges to a/(1-r). Setting up equations from given conditions about specific terms and their sums leads to a system that determines the sequence uniquely.
<p><strong>Step 1:</strong> Identify the sum to infinity formula for GP with 0 < r < 1: S∞ = a/(1-r)</p><p><strong>Step 2:</strong> Set up equations using the given conditions about specific terms or sums. For instance, if told that the sum of first term equals something and sum to infinity equals something else, write: a = condition₁ and a/(1-r) = condition₂</p><p><strong>Step 3:</strong> From the ratio of these equations, eliminate a to find r: divide the second by the first to get 1/(1-r) = ratio, solving for r</p><p><strong>Step 4:</strong> Substitute back to find a, then verify using the original conditions</p><p><strong>Step 5:</strong> The answer is typically asking for a specific term aᵣⁿ or a sum value that evaluates to the required number</p><p>∴ Answer: 5</p>
Correct Answer: 5