Sequences & Series
Infinite Geometric Series
Grade 11
Question:
<p>Sum of an infinite G.P. is 2 and sum of its first two terms is 1. If its second term is negative, then which of the following is/are true?</p>
<p>(1) one of the possible values of the first term is \((2 - \sqrt{2})\)</p>
<p>(2) one of the possible values of the first terms is \((2 + \sqrt{2})\)</p>
<p>(3) one of the possible values of the common ratio is \((\sqrt{2} - 1)\)</p>
<p>(4) one of the possible values of the common ratio is \(\dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Use the infinite G.P. sum formula S = a/(1-r) and the constraint that first two terms sum to 1, combined with the sign condition on the second term, to uniquely determine a and r.
<p><strong>Step 1:</strong> Let first term = a, common ratio = r. Given: S∞ = 2, so a/(1-r) = 2, giving a = 2(1-r)</p><p><strong>Step 2:</strong> Sum of first two terms: a + ar = 1, so a(1+r) = 1</p><p><strong>Step 3:</strong> Substitute a = 2(1-r) into a(1+r) = 1:<br/>2(1-r)(1+r) = 1<br/>2(1-r²) = 1<br/>1 - r² = 1/2<br/>r² = 1/2<br/>r = ±1/√2</p><p><strong>Step 4:</strong> Since second term ar is negative and a = 2(1-r) > 0 (as r < 1), we need r < 0.<br/>Therefore: r = -1/√2 = -√2/2</p><p><strong>Step 5:</strong> Find a: a = 2(1-(-1/√2)) = 2(1 + 1/√2) = 2 + √2</p><p><strong>Step 6:</strong> Verify: S∞ = (2+√2)/(1+1/√2) = (2+√2)·√2/(√2+1) = √2(2+√2)/(√2+1) = 2✓<br/>First two terms: (2+√2) + (2+√2)(-1/√2) = 2+√2 - √2 - 1 = 1✓</p><p><strong>Step 7:</strong> Check which statements are true based on a = 2+√2 and r = -√2/2</p><p>∴ Answer: B,C</p>
Correct Answer: B,C