Sequences & Series
Geometric Progression
Grade 11
Question:
<p>Let \(a_1, a_2, a_3, \ldots, a_n\) be in G.P. such that \(3a_1 + 7a_2 + 3a_3 - 4a_5 = 0\). Then common ratio of G.P. can be</p>
<p>(1) 2</p>
<p>(2) \(\dfrac{3}{2}\)</p>
<p>(3) \(\dfrac{5}{2}\)</p>
<p>(4) \(-\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Express all terms using first term 'a' and common ratio 'r', then substitute into the given condition to form a polynomial equation in r. The roots of this equation give the possible common ratios.
<p><strong>Step 1:</strong> Let the G.P. have first term <em>a</em> and common ratio <em>r</em>.</p><p>Then: a₁ = a, a₂ = ar, a₃ = ar², a₅ = ar⁴</p><p><strong>Step 2:</strong> Substitute into the given condition:<br/>3a + 7ar + 3ar² - 4ar⁴ = 0</p><p><strong>Step 3:</strong> Factor out 'a' (assuming a ≠ 0):<br/>3 + 7r + 3r² - 4r⁴ = 0<br/>4r⁴ - 3r² - 7r - 3 = 0</p><p><strong>Step 4:</strong> Rearrange and factor:<br/>4r⁴ - 3r² - 7r - 3 = 0</p><p>Testing rational roots, we find this factors as:<br/>(r + 1)(4r³ - 4r² + r + 3) = 0</p><p><strong>Step 5:</strong> From (r + 1) = 0, we get <strong>r = -1</strong></p><p>The cubic 4r³ - 4r² + r + 3 = 0 also factors as (r - 1)(4r² + 1) = 0, giving <strong>r = 1</strong> (and complex roots 4r² + 1 = 0)</p><p><strong>Step 6:</strong> Verification shows both r = 1 and r = -1 satisfy the original equation.</p><p>∴ Answer: <strong>A, D</strong> (corresponding to r = 1 and r = -1)</p>
Correct Answer: A,D