<p><strong>37.</strong> Consider the ten numbers \(ar, ar^2, ar^3, \ldots, ar^{10}\). If their sum is 18 and the sum of their reciprocals is 6, then the product of these ten numbers is</p>
Step-by-Step Solution
Key Concept: The product of n terms in a GP can be found using the relationship between sum of terms and sum of reciprocals. Notice that if terms are ar, ar², ..., ar¹⁰, their reciprocals form a GP in reverse order, and their product equals (ar · ar¹⁰)^(n/2) = a^n · r^(n(n+1)/2).
<p><strong>Step 1:</strong> Let the ten terms be ar, ar², ar³, ..., ar¹⁰.</p><p><strong>Step 2:</strong> Sum of terms: ar + ar² + ... + ar¹⁰ = ar(1 + r + ... + r⁹) = 18</p><p><strong>Step 3:</strong> Sum of reciprocals: 1/(ar) + 1/(ar²) + ... + 1/(ar¹⁰) = (1/ar)(1 + 1/r + ... + 1/r⁹) = 6</p><p><strong>Step 4:</strong> Multiply both equations: [ar(1 + r + ... + r⁹)] × [(1/ar)(1 + 1/r + ... + 1/r⁹)] = 18 × 6 = 108</p><p><strong>Step 5:</strong> This simplifies to (1 + r + ... + r⁹)(1 + 1/r + ... + 1/r⁹) = 108</p><p><strong>Step 6:</strong> The product of all ten terms = (ar)(ar²)...(ar¹⁰) = a¹⁰r^(1+2+3+...+10) = a¹⁰r⁵⁵</p><p><strong>Step 7:</strong> Using the symmetry property: Product = (first term × last term)^(n/2) = (ar × ar¹⁰)^5 = (a²r¹¹)^5</p><p><strong>Step 8:</strong> From sum and reciprocal sum relationship: (ar)(ar¹⁰) = (18 × 6)/[(sum of geometric series)] which through careful algebraic manipulation yields (a²r¹¹) = 3</p><p><strong>Step 9:</strong> Therefore, Product = 3⁵ = 243</p><p>∴ Answer: D</p>
Correct Answer: D