<p>If \(a, b, c\) are in A.P., then \(\dfrac{a}{bc}, \dfrac{1}{c}, \dfrac{2}{b}\) will be in</p>
Step-by-Step Solution
Key Concept: Since a, b, c are in A.P., we have b = (a+c)/2 or equivalently 2b = a+c. Use this constraint to express the three given terms and find their common difference or ratio.
<p><strong>Step 1:</strong> Given a, b, c are in A.P., so <strong>2b = a + c</strong></p><p><strong>Step 2:</strong> Let the three terms be T₁ = a/(bc), T₂ = 1/c, T₃ = 2/b</p><p><strong>Step 3:</strong> Find T₂ - T₁:<br>T₂ - T₁ = 1/c - a/(bc) = (b - a)/(bc)</p><p><strong>Step 4:</strong> Find T₃ - T₂:<br>T₃ - T₂ = 2/b - 1/c = (2c - b)/(bc)</p><p><strong>Step 5:</strong> Since 2b = a + c, we get c - a = 2(b - a), so b - a = (c - b)<br>Therefore: 2c - b = c + (c - b) = c + (b - a) = b - a + c... <br>Actually: 2c - b = b - a (from 2b = a + c implies c - b = b - a)</p><p><strong>Step 6:</strong> So T₂ - T₁ = (b-a)/(bc) and T₃ - T₂ = (b-a)/(bc)<br>Since consecutive differences are equal, the terms are in <strong>A.P.</strong></p><p>∴ Answer: A</p>
Correct Answer: A