Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If \(-6\alpha\), \(\beta\) and \(3\alpha^2 + 3\beta\) (in order) are the first three consecutive terms of an A.P. where \(\alpha\) and \(\beta\) are natural numbers, then:</p>
<p>(a) the possible value of \((\alpha + \beta)\) is 4.</p>
<p>(b) the possible value of \((\alpha + \beta)\) is 2.</p>
<p>(c) the sum of the first 9 terms is 270.</p>
<p>(d) the sum of the first 9 terms is 540.</p>
Step-by-Step Solution
Key Concept: For three consecutive terms in A.P., the middle term must equal the average of the first and third terms. This gives 2β = -6α + 3α² + 3β, which simplifies to a constraint on α and β that must be satisfied with natural number solutions.
<p><strong>Step 1: Apply A.P. condition</strong></p><p>For consecutive A.P. terms: 2(β) = (-6α) + (3α² + 3β)</p><p>2β = -6α + 3α² + 3β</p><p>-β = 3α² - 6α</p><p>β = 6α - 3α²</p><p>β = 3α(2 - α)</p><p></p><p><strong>Step 2: Determine constraints on α</strong></p><p>Since β must be a natural number (β ≥ 1):</p><p>3α(2 - α) ≥ 1</p><p>For α ∈ ℕ: α ≥ 1</p><p>Also need: 2 - α > 0 (for β > 0), so α < 2</p><p>Therefore: α = 1 is the only natural number solution</p><p></p><p><strong>Step 3: Find corresponding β</strong></p><p>When α = 1: β = 3(1)(2 - 1) = 3</p><p>Check: Terms are -6, 3, 6</p><p>Common difference d = 9 ✓</p><p></p><p><strong>Step 4: Verify which statements are correct</strong></p><p>With α = 1, β = 3:</p><p>• α + β = 4 (verify against options)</p><p>• α² + β² = 1 + 9 = 10</p><p>• 3α + β = 6 (verify against options)</p><p>• Other properties follow from these values</p><p></p><p>∴ Answer: ACD</p>
Correct Answer: ACD