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: Group consecutive pairs using difference of squares factorization (a² - b²) = (a+b)(a-b), then exploit the arithmetic progression property that consecutive terms have constant difference.
<p><strong>Step 1:</strong> Let the common difference be d. Then aₙ = a₁ + (n-1)d.</p><p><strong>Step 2:</strong> Group consecutive pairs and apply difference of squares:</p><p>S = (a₁² - a₂²) + (a₃² - a₄²) + ... + (a₂ₖ₋₁² - a₂ₖ²)</p><p><strong>Step 3:</strong> Factor each pair using a² - b² = (a+b)(a-b):</p><p>S = (a₁ + a₂)(a₁ - a₂) + (a₃ + a₄)(a₃ - a₄) + ... + (a₂ₖ₋₁ + a₂ₖ)(a₂ₖ₋₁ - a₂ₖ)</p><p><strong>Step 4:</strong> In an AP, (aᵢ - aⱼ) = (j-i)d. Therefore:</p><p>(a₁ - a₂) = -d, (a₃ - a₄) = -d, ..., (a₂ₖ₋₁ - a₂ₖ) = -d</p><p><strong>Step 5:</strong> Factor out -d:</p><p>S = -d[(a₁ + a₂) + (a₃ + a₄) + ... + (a₂ₖ₋₁ + a₂ₖ)]</p><p><strong>Step 6:</strong> The sum in brackets equals sum of all 2k terms = (2k/2)[2a₁ + (2k-1)d] = k[2a₁ + (2k-1)d]</p><p><strong>Step 7:</strong> Therefore:</p><p>S = -kd[2a₁ + (2k-1)d] = -k(a₁ + a₂ₖ)d</p><p>Or equivalently: <strong>S = -kd(a₁ + a₂ₖ)</strong></p><p>∴ Answer: A</p>
Correct Answer: A