Sequences & Series
AM, GM, HM Relations
Grade 11
Question:
<p>316. If the first and \((2n-1)\)th terms of an AP, a GP and an HP are equal and their \(n\)th terms are \(a\), \(b\) and \(c\) respectively, then</p>
<p>(a) \(a = b = c\)</p>
<p>(b) \(a \geq b \geq c\)</p>
<p>(c) \(a + c = b\)</p>
<p>(d) \(ac - b^2 = 0\)</p>
Step-by-Step Solution
Key Concept: Since all three progressions share equal first and (2n-1)th terms, use the property that the nth term is the arithmetic mean of equidistant terms. This means the nth term must be the geometric mean of the 1st and (2n-1)th terms for each progression type.
<p><strong>Step 1:</strong> Let the first term be α and the (2n-1)th term be β. Given: all three progressions have same α and β.</p><p><strong>Step 2:</strong> For AP: The nth term is the arithmetic mean of equidistant terms. So a = (α + β)/2</p><p><strong>Step 3:</strong> For GP: The nth term is the geometric mean of equidistant terms. So b² = α·β, giving b = √(αβ)</p><p><strong>Step 4:</strong> For HP: Terms are reciprocals of an AP. If nth term of HP is c, then 1/c is the arithmetic mean of 1/α and 1/β. So 1/c = (1/α + 1/β)/2, which gives c = 2αβ/(α + β)</p><p><strong>Step 5:</strong> Observe that: a = (α + β)/2 (AM), b = √(αβ) (GM), c = 2αβ/(α + β) (HM)</p><p><strong>Step 6:</strong> The relationship is: <strong>a ≥ b ≥ c</strong> (with equality only when α = β)</p><p><strong>Key Result:</strong> For positive quantities, AM ≥ GM ≥ HM. Therefore: <strong>a ≥ b ≥ c</strong></p><p>∴ Answer: B</p>
Correct Answer: B