Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>Let <i>S</i><sub>1</sub> be the sum of first 2<i>n</i> terms of an arithmetic progression. Let <i>S</i><sub>2</sub> be the sum of first 4<i>n</i> terms of the same arithmetic progression. If (<i>S</i><sub>2</sub> − <i>S</i><sub>1</sub>) is 1000, then the sum of the first 6<i>n</i> terms of the arithmetic progression is equal to</p>
<p>(a) 1000</p>
<p>(b) 7000</p>
<p>(c) 5000</p>
<p>(d) 3000</p>
Step-by-Step Solution
Key Concept: For an AP, the sum of first n terms is Sₙ = n/2[2a + (n-1)d]. The sum of terms from (2n+1) to 4n equals S₂ - S₁, which can be expressed as the sum of 2n consecutive terms of a related AP. Use this relationship to find the common difference pattern.
<p><strong>Step 1:</strong> Set up the sum formulas for an AP with first term a and common difference d.</p><p>S₁ (sum of first 2n terms) = (2n)/2[2a + (2n-1)d] = n[2a + (2n-1)d]</p><p>S₂ (sum of first 4n terms) = (4n)/2[2a + (4n-1)d] = 2n[2a + (4n-1)d]</p><p><strong>Step 2:</strong> Calculate S₂ - S₁.</p><p>S₂ - S₁ = 2n[2a + (4n-1)d] - n[2a + (2n-1)d]</p><p>= 2n(2a) + 2n(4n-1)d - n(2a) - n(2n-1)d</p><p>= n(2a) + n[2(4n-1) - (2n-1)]d</p><p>= n(2a) + n[8n - 2 - 2n + 1]d</p><p>= n(2a) + n(6n - 1)d = 1000</p><p><strong>Step 3:</strong> Find S₃ (sum of first 6n terms).</p><p>S₃ = (6n)/2[2a + (6n-1)d] = 3n[2a + (6n-1)d]</p><p>= 3n(2a) + 3n(6n-1)d</p><p>= 3[n(2a) + n(6n-1)d]</p><p>= 3 × 1000 = 3000</p><p><strong>Step 4:</strong> Verify the pattern. Notice that:</p><p>S₂ - S₁ = n(2a) + n(6n-1)d is exactly the coefficient that appears in S₃</p><p>Therefore: S₃ = 3(S₂ - S₁) = 3 × 1000 = 3000</p><p>∴ Answer: b</p>
Correct Answer: b