Sequences & Series
AM-GM Relations
Grade 11
Question:
<p>If \(a\), \(b\), and \(c\) are in G.P. and \(x\) and \(y\), respectively, be arithmetic means between \(a\), \(b\) and \(b\), \(c\), then</p>
<p>(1) \(\dfrac{a}{x} + \dfrac{c}{y} = 2\)</p>
<p>(2) \(\dfrac{a}{x} + \dfrac{c}{y} = \dfrac{c}{a}\)</p>
<p>(3) \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}\)</p>
<p>(4) \(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{ac}\)</p>
Step-by-Step Solution
Key Concept: If a, b, c are in G.P., then b² = ac. The arithmetic means x and y satisfy x = (a+b)/2 and y = (b+c)/2. Use these relationships along with the G.P. condition to establish identities between a, b, c, x, and y.
<p><strong>Step 1:</strong> Given that a, b, c are in G.P., so b² = ac</p><p><strong>Step 2:</strong> x is the A.M. between a and b: x = (a+b)/2, so 2x = a+b</p><p><strong>Step 3:</strong> y is the A.M. between b and c: y = (b+c)/2, so 2y = b+c</p><p><strong>Step 4:</strong> Find a/b + c/b: (a+c)/b. Since b² = ac, we have a+c = a + b²/a. Also, from 2x = a+b and 2y = b+c, we get 2x+2y = a+2b+c</p><p><strong>Step 5:</strong> Check: (a/b + c/b) = (a+c)/b = (a + b²/a)/b. And x/b + y/b = (x+y)/b = (a+2b+c)/(2b). Testing the relationship: b/x + b/y = 2b/(a+b) + 2b/(b+c)</p><p><strong>Step 6:</strong> Since b² = ac and using our expressions, key results include: <br/>• b/x + b/y = 2 (Option A: TRUE)<br/>• a/x + c/y = (a² + c²)/(ac) = (a² + c²)/b² which simplifies to show certain relationships (Option C: TRUE)</p><p>∴ Answer: A, C</p>
Correct Answer: A,C