<p>The sum of an infinite G.P. is 57 and the sum of their cubes is 9747, then the common ratio of the G.P. is</p>
Step-by-Step Solution
Key Concept: For an infinite G.P. with first term a and common ratio r (|r| < 1), if S = a/(1-r) and the sum of cubes forms another G.P. with first term a³ and common ratio r³, then S³ = a³/(1-r³). Use these two relations to find r.
<p><strong>Step 1:</strong> Set up equations for the given conditions.</p><p>Let first term = a, common ratio = r where |r| < 1</p><p>Sum of G.P.: S = a/(1-r) = 57 ... (1)</p><p>Sum of cubes: a³ + a³r³ + a³r⁶ + ... = a³/(1-r³) = 9747 ... (2)</p><p><strong>Step 2:</strong> Use the factorization 1 - r³ = (1-r)(1+r+r²)</p><p>From (2): a³/[(1-r)(1+r+r²)] = 9747</p><p>a³/(1-r) · 1/(1+r+r²) = 9747</p><p><strong>Step 3:</strong> From (1), a/(1-r) = 57, so a³/(1-r) = 57³a²</p><p>Therefore: 57a² · 1/(1+r+r²) = 9747</p><p>a²/(1+r+r²) = 171</p><p><strong>Step 4:</strong> From (1): a = 57(1-r)</p><p>Substituting: [57(1-r)]²/(1+r+r²) = 171</p><p>57²(1-r)²/(1+r+r²) = 171</p><p>3249(1-r)²/(1+r+r²) = 171</p><p>(1-r)²/(1+r+r²) = 171/3249 = 1/19</p><p><strong>Step 5:</strong> Cross multiply: 19(1-r)² = 1+r+r²</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>(2r-3)(3r-2) = 0</p><p>r = 3/2 or r = 2/3</p><p><strong>Step 6:</strong> Since |r| < 1 for convergence, r = 2/3</p><p>∴ Answer: B</p>
Correct Answer: B