<p>Which term of the sequence \( 2, 1, 2^{-1}, 4^{-1}, 8^{-1}, \ldots \) is \( \dfrac{1}{128} \)?</p>
Step-by-Step Solution
Key Concept: Recognize this as a geometric sequence after the first term, then identify which power of 2 equals 1/128 by expressing it as 2^(-7).
<p><strong>Step 1:</strong> Identify the pattern. After the first term, we have: 1, 2^(-1), 4^(-1), 8^(-1), ... = 1, 2^(-1), 2^(-2), 2^(-3), ...</p><p><strong>Step 2:</strong> The 2nd term onwards forms a GP with first term a = 1 and common ratio r = 1/2 = 2^(-1).</p><p><strong>Step 3:</strong> The nth term of the full sequence is: T_n = 2^(-(n-2)) for n ≥ 2.</p><p><strong>Step 4:</strong> Set T_n = 1/128. Since 1/128 = 2^(-7), we have: 2^(-(n-2)) = 2^(-7)</p><p><strong>Step 5:</strong> Therefore -(n-2) = -7, which gives n - 2 = 7, so n = 9.</p><p><strong>Verification:</strong> T_9 = 2^(-(9-2)) = 2^(-7) = 1/128 ✓</p><p>∴ Answer: <strong>9th term</strong> (Option C)</p>
Correct Answer: C