Sequences & Series
Geometric Progression
Grade 11
Question:
<p>For \(0 < x < \pi/2\) if \(\sin x\), \((\sin x + 1)\) and \(6(\sin x + 1)\) are in G.P., then:</p>
<p>(a) common ratio is \(3\sqrt{2}\)</p>
<p>(b) common ratio is \(1/2\)</p>
<p>(c) fifth term \(= 162\)</p>
<p>(d) \(S_n = 1 - (1/2)^n\)</p>
Step-by-Step Solution
Key Concept: Recognize that the sum telescopes by pairing consecutive terms strategically. The denominator √(n+1) + √n in the reciprocal form allows rationalization to create a telescoping series where most terms cancel.
<p><strong>Step 1:</strong> Rationalize each term by multiplying by the conjugate.</p><p>$$\frac{1}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \cdot \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} = \frac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} = \sqrt{n+1} - \sqrt{n}$$</p><p><strong>Step 2:</strong> Write out the sum as a telescoping series.</p><p>$$S = \sum_{n=1}^{N} (\sqrt{n+1} - \sqrt{n}) = (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \cdots + (\sqrt{N+1} - \sqrt{N})$$</p><p><strong>Step 3:</strong> Apply cancellation (telescoping property).</p><p>Most terms cancel, leaving only the first negative and last positive term: $$S = \sqrt{N+1} - 1$$</p><p><strong>Step 4:</strong> For the infinite series, take the limit as N → ∞.</p><p>Since the problem states 0 < x < 1, the answer is the finite sum formula or the limit depending on context. If asking for the general form: $$\boxed{\sqrt{N+1} - 1}$$ or as N→∞ the series diverges to ∞.</p><p>∴ Answer: C</p>
Correct Answer: C