Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(a_1, a_2, a_3, \ldots\) are in arithmetic progression, then \(S = a_1^2 - a_2^2 + a_3^2 - a_4^2 + \ldots - a_{2k}^2\) is equal to:</p>
<p>(a) \(\dfrac{k}{2k-1}(a_1^2 - a_{2k}^2)\)</p>
<p>(b) \(\dfrac{2k}{k-1}(a_{2k}^2 - a_1^2)\)</p>
<p>(c) \(\dfrac{k}{k+1}(a_1^2 - a_{2k}^2)\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use the difference of squares formula a² - b² = (a+b)(a-b) repeatedly, then exploit the arithmetic progression property to find a telescoping or factored pattern.
<p><strong>Step 1:</strong> Group consecutive pairs using difference of squares:</p><p>S = (a₁² - a₂²) + (a₃² - a₄²) + ... + (a₂ₖ₋₁² - a₂ₖ²)</p><p><strong>Step 2:</strong> Apply difference of squares formula to each pair:</p><p>S = (a₁ - a₂)(a₁ + a₂) + (a₃ - a₄)(a₃ + a₄) + ... + (a₂ₖ₋₁ - a₂ₖ)(a₂ₖ₋₁ + a₂ₖ)</p><p><strong>Step 3:</strong> In an AP with common difference d: aᵢ - aᵢ₊₁ = -d for all i</p><p>S = -d(a₁ + a₂) - d(a₃ + a₄) - ... - d(a₂ₖ₋₁ + a₂ₖ)</p><p>S = -d[(a₁ + a₂) + (a₃ + a₄) + ... + (a₂ₖ₋₁ + a₂ₖ)]</p><p><strong>Step 4:</strong> Sum inside brackets has k pairs. Using AP property aᵢ + aᵢ₊₁ = 2aᵢ + d:</p><p>S = -d · k(a₁ + a₂ₖ) or equivalently S = -kd(2a₁ + (2k-1)d)</p><p>∴ Answer: A</p>
Correct Answer: A