Sequences & Series
Arithmetic Mean
Grade 11
Question:
<p>A sequence \(a_1, a_2, a_3, \ldots, a_n\) of real numbers is such that \(a_1 = 0\), \(|a_2| = |a_1 + 1|\), \(|a_3| = |a_2 + 1|\), \(\ldots\), \(|a_n| = |a_{n-1} + 1|\), where the arithmetic mean of \(a_1, a_2, a_3, \ldots, a_n\) cannot be less than \(-\lambda/\mu\), then find the value of \(\lambda + \mu\)</p>
<p>\(\lambda + \mu = 3\)</p>
<p>\(\lambda\mu = 2\)</p>
<p>\(\lambda^\mu + \mu^\lambda = 3\)</p>
<p>\(\lambda\) & \(\mu\) can be roots of a quadratic equation with rational co-efficient</p>
Step-by-Step Solution
Key Concept: At each step, |a_n| = |a_{n-1} + 1| creates two branches (positive and negative); to minimize the arithmetic mean, we must optimally choose signs at each step. The minimum mean occurs when we greedily choose the most negative value at each stage, which follows the pattern a_n = -(a_{n-1} + 1) when possible.
<p><strong>Step 1: Analyze the recurrence relation</strong></p><p>Given: a₁ = 0, and |aₙ| = |aₙ₋₁ + 1| for n ≥ 2. At each step, aₙ = ±|aₙ₋₁ + 1|.</p><p><strong>Step 2: Find the branch that minimizes the sum</strong></p><p>Starting with a₁ = 0:<br/>• a₂: |a₂| = |0 + 1| = 1, so a₂ ∈ {1, -1}. Choose a₂ = -1 (to minimize)<br/>• a₃: |a₃| = |-1 + 1| = 0, so a₃ = 0<br/>• a₄: |a₄| = |0 + 1| = 1, so a₄ ∈ {1, -1}. Choose a₄ = -1<br/>• a₅: |a₅| = |-1 + 1| = 0, so a₅ = 0<br/>Pattern repeats: 0, -1, 0, -1, 0, -1, ...</p><p><strong>Step 3: Calculate the arithmetic mean</strong></p><p>For n terms following the pattern (0, -1) repeating:<br/>• If n = 2k: Sum = k(0 - 1) = -k, Mean = -k/(2k) = -1/2<br/>• If n = 2k+1: Sum = k(0 - 1) + 0 = -k, Mean = -k/(2k+1) → -1/2 as n → ∞<br/>The minimum value the mean can approach is -1/2.</p><p><strong>Step 4: Express in the form -λ/μ</strong></p><p>The arithmetic mean cannot be less than -1/2, so λ/μ = 1/2 where λ = 1, μ = 2 (in lowest terms).</p><p>∴ λ + μ = 1 + 2 = <strong>3</strong></p>
Correct Answer: A