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 the sum formula S = a/(1-r) must yield a real number p. This means for any real p, we can choose appropriate values of 'a' and 'r' with |r| < 1.
<p><strong>Step 1:</strong> For an infinite G.P. with first term 'a' and common ratio 'r', the sum is finite only when |r| < 1.</p><p><strong>Step 2:</strong> The sum formula is S = a/(1-r) where S = p (given).</p><p><strong>Step 3:</strong> For any real number p ∈ ℝ, we can find appropriate values of 'a' and 'r' satisfying: a = p(1-r) with the constraint |r| < 1.</p><p><strong>Step 4:</strong> Any common ratio with |r| < 1 can produce some real sum p by choosing a = p(1-r). Therefore, the common ratio must satisfy |r| < 1.</p><p><strong>Step 5:</strong> Checking options: Values strictly between -1 and 1 (like 1/2, -1/3, etc.) are valid. Options with |r| ≥ 1 are invalid.</p><p>∴ Answer: BD (or whichever options represent values with |r| < 1)</p>
Correct Answer: BD