Sequences & Series
Infinite Geometric Series
Grade 11
Question:
<p>Given sum of an infinite geometric series with positive terms is <strong>3</strong> and the sum of the cubes of its terms is \(\dfrac{27}{19}\). Find the common ratio of the G.P.</p>
<p>\(r = \dfrac{1}{3}\)</p>
<p>\(r = \dfrac{2}{3}\)</p>
<p>\(r = \dfrac{1}{2}\)</p>
<p>\(r = \dfrac{3}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the formula for sum of infinite G.P. and recognize that cubing each term of a G.P. with first term 'a' and ratio 'r' creates a new G.P. with first term 'a³' and ratio 'r³'. Set up two equations from the given conditions.
<p><strong>Step 1:</strong> Let first term = a, common ratio = r, where |r| < 1.</p><p>Given: Sum of series = a/(1-r) = 3, so <strong>a = 3(1-r)</strong></p><p><strong>Step 2:</strong> The cubes of terms form a G.P.: a³, a³r³, a³r⁶, ... with first term a³ and common ratio r³.</p><p>Sum of cubes = a³/(1-r³) = 27/19</p><p><strong>Step 3:</strong> Note that 1 - r³ = (1-r)(1 + r + r²). So:</p><p>a³/[(1-r)(1 + r + r²)] = 27/19</p><p><strong>Step 4:</strong> Substitute a = 3(1-r):</p><p>[3(1-r)]³/[(1-r)(1 + r + r²)] = 27/19</p><p>27(1-r)³/[(1-r)(1 + r + r²)] = 27/19</p><p>27(1-r)²/(1 + r + r²) = 27/19</p><p><strong>Step 5:</strong> Simplify:</p><p>(1-r)²/(1 + r + r²) = 1/19</p><p>19(1 - 2r + r²) = 1 + r + r²</p><p>19 - 38r + 19r² = 1 + r + r²</p><p>18r² - 39r + 18 = 0</p><p>6r² - 13r + 6 = 0</p><p><strong>Step 6:</strong> Using quadratic formula: r = (13 ± √(169-144))/12 = (13 ± 5)/12</p><p>r = 3/2 or r = 2/3</p><p><strong>Step 7:</strong> Since |r| < 1 for convergence, r = 2/3</p><p>∴ Answer: B (r = 2/3)</p>
Correct Answer: B