<p>The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes of its terms is \(\dfrac{27}{19}\). Then the common ratio of this series is:</p>
Step-by-Step Solution
Key Concept: For a GP with first term 'a' and common ratio 'r', the series of cubes forms another GP with first term 'a³' and common ratio 'r³'. Set up two equations using the sum formulas and solve simultaneously.
<p><strong>Step 1:</strong> Let the GP have first term <em>a</em> and common ratio <em>r</em> (where 0 < <em>r</em> < 1 for convergence).</p><p>Sum of series: $\frac{a}{1-r} = 3$ ... (1)</p><p><strong>Step 2:</strong> The cubes of terms form a new GP: $a^3, a^3r^3, a^3r^6, ...$ with first term $a^3$ and common ratio $r^3$.</p><p>Sum of cubes: $\frac{a^3}{1-r^3} = \frac{27}{19}$ ... (2)</p><p><strong>Step 3:</strong> From equation (1): $a = 3(1-r)$</p><p><strong>Step 4:</strong> Substitute into equation (2):</p><p>$\frac{[3(1-r)]^3}{1-r^3} = \frac{27}{19}$</p><p>$\frac{27(1-r)^3}{1-r^3} = \frac{27}{19}$</p><p>$\frac{(1-r)^3}{(1-r)(1+r+r^2)} = \frac{1}{19}$</p><p><strong>Step 5:</strong> Simplify:</p><p>$\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$</p><p>$19(1-2r+r^2) = 1+r+r^2$</p><p>$19-38r+19r^2 = 1+r+r^2$</p><p>$18r^2-39r+18 = 0$</p><p>$6r^2-13r+6 = 0$</p><p><strong>Step 6:</strong> Using quadratic formula or factoring: $(2r-3)(3r-2) = 0$</p><p>$r = \frac{3}{2}$ or $r = \frac{2}{3}$</p><p>Since $0 < r < 1$ for convergence: $r = \frac{2}{3}$</p><p>∴ Answer: D</p>
Correct Answer: D