Sequences & Series
AM-GM Inequality in AP
Grade 11
Question:
<p>If \(a_1, a_2, a_3, \ldots, a_n\) are positive numbers in AP whose common difference is \(d\) and \(s_n = \displaystyle\sum_{k=1}^{n} a_k\), then</p>
<p>A. \(s_n \geq n \cdot \sqrt{a_1^2 + (n-1)da_1}\)</p>
<p>B. \(s_n \geq n \cdot \sqrt[n]{a_1 \cdot a_2 \cdots a_n}\)</p>
<p>C. \(\displaystyle\sum_{k=1}^{n} \dfrac{1}{a_k} \cdot \sqrt[n]{a_1 \cdot a_2 \cdots a_n} \geq n\)</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: For an AP with first term a₁ and common difference d, the sum formula sₙ = n/2[2a₁ + (n-1)d] and the reciprocal sum 1/s₁ + 1/s₂ + ... + 1/sₙ can be decomposed using telescoping series when expressed as differences of reciprocals.
<p><strong>Step 1:</strong> Write the sum formula. For AP: sₙ = n/2[2a₁ + (n-1)d]</p><p><strong>Step 2:</strong> Note that sₖ - sₖ₋₁ = aₖ. Since all terms are positive and in AP with difference d > 0, we have: sₖ = sₖ₋₁ + a₁ + (k-1)d</p><p><strong>Step 3:</strong> Express 1/sₖ using telescoping. Multiply and divide: 1/aₖ = 1/d · [1/sₖ₋₁ - 1/sₖ] (This requires verifying that sₖ - sₖ₋₁ = d·sₖ·sₖ₋₁/[sₖ₋₁ + constant])</p><p><strong>Step 4:</strong> For the standard form ∑(1/sₖ) from k=1 to n, use: 1/sₖ = (1/d)[1/sₖ₋₁ - 1/sₖ] when the AP satisfies specific conditions</p><p><strong>Step 5:</strong> The telescoping sum gives: ∑₁ⁿ (1/sₖ) = (1/d)[1/s₀ - 1/sₙ] or equivalently (1/d)[1/a₁ - 1/sₙ]</p><p><strong>Step 6:</strong> Simplify to obtain: ∑₁ⁿ (1/sₖ) = (1/d) · (sₙ - a₁)/(a₁·sₙ) = (n/[2a₁ + (n-1)d])·(1/d)</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B