<p>The numbers 1, 4, 16 can be three terms (not necessarily consecutive) of</p>
Step-by-Step Solution
Key Concept: A geometric sequence is uniquely determined by any two of its terms; once you know two terms and their positions, you can find the common ratio and check if other numbers can be part of that sequence. Test if 1, 4, 16 can each belong to the same geometric progression by verifying the common ratio remains consistent.
<p><strong>Step 1:</strong> Let 1, 4, 16 be the pth, qth, and rth terms respectively of a GP with first term 'a' and common ratio 'd'.</p><p><strong>Step 2:</strong> Then: a·d^(p-1) = 1, a·d^(q-1) = 4, a·d^(r-1) = 16</p><p><strong>Step 3:</strong> Dividing consecutive equations: d^(q-p) = 4 and d^(r-q) = 4</p><p><strong>Step 4:</strong> This gives d^(q-p) = d^(r-q), so if d ≠ 1, then q - p = r - q, meaning these terms are equally spaced in the sequence.</p><p><strong>Step 5:</strong> Let q - p = r - q = k. Then: d^k = 4, so d = 4^(1/k) for any positive integer k. Choose k = 1: d = 4, giving us 1, 4, 16 as consecutive terms. For any k, we can find corresponding 'a' and verify the sequence exists.</p><p><strong>Step 6:</strong> The three numbers can be terms of infinitely many geometric progressions with different common ratios (like 2, 8, 32,... or √2, 4, 8√2,...).</p><p>∴ Answer: C</p>
Correct Answer: C