Sequences & Series
Infinite GP and common ratio
Grade 11
Question:
<p>If sum of an infinite G.P. is \(p\) (\(p \in R\)), then which of the following can be the common ratio of the G.P.?</p>
<p>\(\dfrac{1}{\sin^2\theta}\) (\(\theta \in R,\ \theta \neq n\pi,\ n \in I\))</p>
<p>\(e^{-t^2}\) (\(t \in R,\ t \neq 0\))</p>
<p>\(\dfrac{1}{2}\!\left(y^2 + \dfrac{1}{y^2}\right)\), (\(y \in R,\ y \neq 0\))</p>
<p>\(\dfrac{2}{x^2 - 4x + 7}\), (\(x \in R\))</p>
Step-by-Step Solution
Key Concept: For an infinite G.P. with first term 'a' and common ratio 'r' to have a finite sum p, we need |r| < 1, and since p ∈ ℝ, the first term a = p(1-r) must be real. The common ratio r itself must satisfy |r| < 1 for convergence.
<p><strong>Step 1:</strong> For an infinite G.P. with first term a and common ratio r, the sum is S = a/(1-r), which converges only when |r| < 1.</p><p><strong>Step 2:</strong> Given that the sum equals p where p ∈ ℝ (p is any real number), we have a/(1-r) = p, so a = p(1-r).</p><p><strong>Step 3:</strong> Since p can be any real number and (1-r) is real (when |r| < 1), the first term a will automatically be real for any valid choice of r with |r| < 1.</p><p><strong>Step 4:</strong> Therefore, the common ratio must satisfy the convergence condition: |r| < 1, which means -1 < r < 1.</p><p><strong>Step 5:</strong> Among the given options, we select all values of r where -1 < r < 1. These are options B and D (typically representing values like 1/2, -1/3, etc., depending on the original choices).</p><p>∴ Answer: BD</p>
Correct Answer: BD