Sequences & Series
Arithmetic and Geometric Progressions
Grade 11
Question:
<p>Let <i>b</i><sub>i</sub> > 1 for <i>i</i> = 1, 2, ..., 101. Suppose log<sub>e</sub><i>b</i><sub>1</sub>, log<sub>e</sub><i>b</i><sub>2</sub>, ..., log<sub>e</sub><i>b</i><sub>101</sub> are in Arithmetic Progression (A.P.) with the common difference log<sub>e</sub>2. Suppose <i>a</i><sub>1</sub>, <i>a</i><sub>2</sub>, ..., <i>a</i><sub>101</sub> are in A.P. such that <i>a</i><sub>1</sub> = <i>b</i><sub>1</sub> and <i>a</i><sub>51</sub> = <i>b</i><sub>51</sub>. If <i>t</i> = <i>b</i><sub>1</sub> + <i>b</i><sub>2</sub> + ... + <i>b</i><sub>51</sub> and <i>s</i> = <i>a</i><sub>1</sub> + <i>a</i><sub>2</sub> + ... + <i>a</i><sub>51</sub>, then</p>
<p>(A) <i>s</i> > <i>t</i> and <i>a</i><sub>101</sub> > <i>b</i><sub>101</sub></p>
<p>(B) <i>s</i> > <i>t</i> and <i>a</i><sub>101</sub> < <i>b</i><sub>101</sub></p>
<p>(C) <i>s</i> < <i>t</i> and <i>a</i><sub>101</sub> > <i>b</i><sub>101</sub></p>
<p>(D) <i>s</i> < <i>t</i> and <i>a</i><sub>101</sub> < <i>b</i><sub>101</sub></p>
Step-by-Step Solution
Key Concept: Geometric sequences with ratio > 1 grow faster than arithmetic sequences in the long term, even if they start and meet at the same points.
<p><strong>Analysis:</strong> The sequence {<i>b</i><sub>i</sub>} is geometric with ratio 2, while {<i>a</i><sub>i</sub>} is arithmetic. Both share the same first and 51st terms. For a geometric sequence with ratio > 1, terms grow exponentially; for an arithmetic sequence, growth is linear. The sum of the first 51 terms of the arithmetic sequence exceeds that of the geometric sequence, but beyond the 51st term, the geometric sequence's exponential growth causes <i>a</i><sub>101</sub> < <i>b</i><sub>101</sub>.</p>
Correct Answer: B