Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If the sum of \(n\) terms of an A.P. is \(cn(n-1)\), where \(c \neq 0\), then the sum of the squares of these terms is</p>
<p>\(c^2 n(n+1)^2\)</p>
<p>\(\dfrac{2}{3} c^2 n(n-1)(2n-1)\)</p>
<p>\(\dfrac{2c^2}{3} n(n+1)(2n+1)\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: First extract the terms using Sₙ = cₙ(n-1), then find individual terms using aₙ = Sₙ - Sₙ₋₁, and finally compute the sum of squares using the arithmetic progression property.
<p><strong>Step 1:</strong> Given Sₙ = cₙ(n-1). Find the first term and common difference.</p><p>S₁ = c(1)(0) = 0, so a₁ = 0</p><p>S₂ = c(2)(1) = 2c, so a₁ + a₂ = 2c → a₂ = 2c</p><p>Common difference d = a₂ - a₁ = 2c</p><p><strong>Step 2:</strong> Find the general term of the A.P.</p><p>aₙ = a₁ + (n-1)d = 0 + (n-1)(2c) = 2c(n-1)</p><p><strong>Step 3:</strong> Verify using Sₙ - Sₙ₋₁.</p><p>aₙ = cₙ(n-1) - c(n-1)(n-2) = c(n-1)[n - (n-2)] = 2c(n-1) ✓</p><p><strong>Step 4:</strong> Calculate sum of squares.</p><p>Σ(aₙ)² = Σ[2c(n-1)]² = 4c² Σ(n-1)²</p><p>= 4c² [0² + 1² + 2² + ... + (n-1)²]</p><p>= 4c² · [(n-1)n(2n-1)/6]</p><p>= <strong>2c²n(n-1)(2n-1)/3</strong></p><p>∴ Answer: B</p>
Correct Answer: B