Sequences & Series
Arithmetic Progression
Grade 11
Question:
<p>If in a triangle \(ABC\), \(a\cos^2\!\left(\dfrac{C}{2}\right) + c\cos^2\!\left(\dfrac{A}{2}\right) = \dfrac{3b}{2}\), then the sides \(a\), \(b\) and \(c\)</p>
<p>are in AP.</p>
<p>are in GP.</p>
<p>are in HP.</p>
<p>satisfy \(a + b = c\).</p>
Step-by-Step Solution
Key Concept: Use the half-angle formula cos²(θ/2) = (1+cosθ)/2 and the cosine rule to convert the given trigonometric condition into an algebraic relation between sides a, b, c. This reveals whether the sides form an AP.
<p><strong>Step 1:</strong> Apply half-angle formula: cos²(θ/2) = (1+cosθ)/2</p><p>a·(1+cos C)/2 + c·(1+cos A)/2 = 3b/2</p><p>a(1+cos C) + c(1+cos A) = 3b</p><p><strong>Step 2:</strong> Use cosine rule: cos A = (b²+c²-a²)/(2bc) and cos C = (a²+b²-c²)/(2ab)</p><p>a + a·(a²+b²-c²)/(2ab) + c + c·(b²+c²-a²)/(2bc) = 3b</p><p>a + (a²+b²-c²)/(2b) + c + (b²+c²-a²)/(2b) = 3b</p><p><strong>Step 3:</strong> Simplify the fractions:</p><p>a + c + [(a²+b²-c²) + (b²+c²-a²)]/(2b) = 3b</p><p>a + c + 2b²/(2b) = 3b</p><p>a + c + b = 3b</p><p>a + c = 2b</p><p><strong>Conclusion:</strong> The sides a, b, c are in <strong>Arithmetic Progression (AP)</strong> with b as the middle term.</p><p>∴ Answer: A</p>
Correct Answer: A