<p>If three positive numbers <i>a</i>, <i>b</i> and <i>c</i> are in AP such that <i>abc</i> = 8, then the minimum possible value of <i>b</i> is</p>
Step-by-Step Solution
Key Concept: If a, b, c are in AP with abc = 8, then a = b - d and c = b + d for some d ≥ 0. Substitute into the product constraint to express it entirely in terms of b and d, then minimize b using calculus or AM-GM inequality.
<p><strong>Step 1:</strong> Let a, b, c be in AP. Then a = b - d and c = b + d where d ≥ 0 (d is the common difference).</p><p><strong>Step 2:</strong> Use the constraint abc = 8:<br/>(b - d)(b)(b + d) = 8<br/>b(b² - d²) = 8</p><p><strong>Step 3:</strong> For fixed b, we need b(b² - d²) = 8, which gives d² = b² - 8/b. For d to be real and non-negative, we require:<br/>b² - 8/b ≥ 0<br/>b³ ≥ 8<br/>b ≥ 2</p><p><strong>Step 4:</strong> The minimum value of b occurs when d² = 0 (i.e., d = 0), which means a = b = c. This gives:<br/>b³ = 8<br/>b = 2</p><p><strong>Step 5:</strong> Verify: When b = 2, we have a = b = c = 2, and abc = 8 ✓. As b increases beyond 2, d² becomes positive, confirming b = 2 is the minimum.</p><p>∴ Answer: <strong>b<sub>min</sub> = 2</strong></p>
Correct Answer: D