Sequences & Series
Harmonic Progression
Grade 11
Question:
<p>If \(x = \frac{a}{1+b}\), \(y = \frac{b}{1+c}\), \(z = \frac{c}{1+a}\) where \(a, b, c\) are in A.P. and \(|a| < 1\), \(|b| < 1\), \(|c| < 1\), then \(x, y, z\) are in</p>
<p>(A) G.P.</p>
<p>(B) A.P.</p>
<p>(C) Arithmetic-Geometric Progression</p>
<p>(D) H.P.</p>
Step-by-Step Solution
Key Concept: If the reciprocals of three numbers are in A.P., then the numbers themselves are in H.P.
<p><strong>Solution:</strong> Since \(a, b, c\) are in A.P., we have \(2b = a + c\)</p><p>Note that \(xyz = \frac{abc}{(1+a)(1+b)(1+c)}\)</p><p>We need to show that \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P.</p><p>\(\frac{1}{x} = \frac{1+b}{a}\), \(\frac{1}{y} = \frac{1+c}{b}\), \(\frac{1}{z} = \frac{1+a}{c}\)</p><p>By the constraint \(2b = a + c\) and algebraic manipulation:</p><p>\(2 \cdot \frac{1+c}{b} = \frac{1+b}{a} + \frac{1+a}{c}\) holds, proving \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P.</p><p>Therefore \(x, y, z\) are in H.P.</p><p>∴ Answer is D.</p>
Correct Answer: D