<p>If \(b_1, b_2, b_3\) (\(b_1 > 0\)) are three successive terms of a G.P. with common ratio \(r\), the value of \(r\) for which the inequality \(b_3 > 4b_2 - 3b_1\) holds is given by</p>
Step-by-Step Solution
Key Concept: Express all terms using the first term and common ratio (b₁, b₁r, b₁r²), then substitute into the inequality to isolate r. Since b₁ > 0, you can divide by it without changing the inequality direction.
<p><strong>Step 1:</strong> Write the three successive G.P. terms using first term b₁ and common ratio r:</p><p>b₁ = b₁, b₂ = b₁r, b₃ = b₁r²</p><p><strong>Step 2:</strong> Substitute into the inequality b₃ > 4b₂ − 3b₁:</p><p>b₁r² > 4b₁r − 3b₁</p><p><strong>Step 3:</strong> Since b₁ > 0, divide throughout by b₁:</p><p>r² > 4r − 3</p><p><strong>Step 4:</strong> Rearrange to standard form:</p><p>r² − 4r + 3 > 0</p><p><strong>Step 5:</strong> Factor the quadratic:</p><p>(r − 1)(r − 3) > 0</p><p><strong>Step 6:</strong> Analyze the sign: The product is positive when both factors have the same sign.</p><p>• Both positive: r > 3</p><p>• Both negative: r < 1</p><p><strong>Step 7:</strong> The solution is r < 1 or r > 3 (r ≠ 0 since it's a G.P.)</p><p>∴ Answer: r ∈ (−∞, 1) ∪ (3, ∞)</p>
Correct Answer: A