<p>The first term of an infinite geometric series is 21. The second term and the sum of the series are both positive integers. Then which of the following is not the possible value of the second term?</p>
Step-by-Step Solution
Key Concept: For a geometric series with first term a=21 and common ratio r, the second term is 21r and sum is 21/(1-r). Both must be positive integers, which constrains the possible values of r through divisibility conditions.
<p><strong>Step 1:</strong> Let the common ratio be r. Second term = 21r (must be a positive integer), and sum S = 21/(1-r) (must be a positive integer).</p><p><strong>Step 2:</strong> For convergence, |r| < 1. Since second term is positive, r > 0, so 0 < r < 1.</p><p><strong>Step 3:</strong> Let second term = 21r = m (positive integer). Then r = m/21.</p><p><strong>Step 4:</strong> Sum = 21/(1 - m/21) = 21/((21-m)/21) = 441/(21-m). For this to be a positive integer, (21-m) must divide 441 = 21² = 3² × 7².</p><p><strong>Step 5:</strong> The divisors of 441 are: 1, 3, 7, 9, 21, 49, 63, 147, 441. Since 0 < r < 1, we need 0 < m < 21, so 21-m must be a divisor of 441 with 21-m > 0.</p><p><strong>Step 6:</strong> Valid values of (21-m): 1, 3, 7, 9, 21, 49, 63, 147, 441 with m < 21 gives (21-m) ∈ {1, 3, 7, 9, 21}.</p><p><strong>Step 7:</strong> This yields m ∈ {20, 18, 14, 12, 0}. Since m > 0, possible second terms are: 20, 18, 14, 12.</p><p>∴ Any value NOT in {12, 14, 18, 20} is not possible for the second term. (The answer C should be such a value, e.g., 15, 16, 17, 19, etc.)</p>
Correct Answer: C