Sequences & Series
Geometric Progression
Grade 11
Question:
<p>The terms of an infinitely decreasing G.P. having common ratio <i>r</i> in which all the terms are positive, the first term is 4, and the difference between the third and fifth terms is 32/81, then</p>
<p>(1) \(r = 1/3\)</p>
<p>(2) \(r = 2\sqrt{2}/3\)</p>
<p>(3) Sum of infinite terms is 6</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: For a decreasing G.P. with first term a and ratio r (0 < r < 1), use the formula for nth term (ar^(n-1)) to set up equations. The constraint that terms are positive and decreasing determines the sign and magnitude of r uniquely.
<p><strong>Step 1:</strong> Set up the G.P. with first term a = 4 and common ratio r. The terms are: 4, 4r, 4r², 4r³, 4r⁴, ...</p><p><strong>Step 2:</strong> Express third and fifth terms: T₃ = 4r² and T₅ = 4r⁴</p><p><strong>Step 3:</strong> Use the given condition T₃ - T₅ = 32/81</p><p>4r² - 4r⁴ = 32/81</p><p>4r²(1 - r²) = 32/81</p><p>r²(1 - r²) = 8/81</p><p><strong>Step 4:</strong> Let r² = x, then x(1 - x) = 8/81</p><p>x - x² = 8/81</p><p>81x² - 81x + 8 = 0</p><p><strong>Step 5:</strong> Using quadratic formula: x = [81 ± √(6561 - 2592)]/162 = [81 ± √3969]/162 = [81 ± 63]/162</p><p>x = 144/162 = 8/9 or x = 18/162 = 1/9</p><p><strong>Step 6:</strong> Since 0 < r < 1 for decreasing series: r² = 8/9 gives r = 2√2/3, or r² = 1/9 gives r = 1/3</p><p>Both values satisfy 0 < r < 1.</p><p><strong>Step 7:</strong> Calculate sum to infinity S = a/(1-r) = 4/(1-r)</p><p>For r = 1/3: S = 4/(2/3) = 6</p><p>For r = 2√2/3: S = 4/(1 - 2√2/3) = 12/(3 - 2√2)</p><p>∴ Answer: A,C (Both r = 1/3 and r = 2√2/3 are valid solutions)</p>
Correct Answer: A,C