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COMEDK · Maths · 5. Sequences and Series

Given \(a, b, c\) are three unequal numbers such that \(\mathrm{b}\) is arithmetic mean of \(a\) and \(c\) and \((b-a),(c-b), a\) are in geometric progression. Then \(a: b: c\) is

  1. A \(1: 3: 5\)
  2. B \(2: 3: 5\)
  3. C \(1: 2: 3\)
  4. D \(1: 2: 4\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1: 2: 3\)

Step-by-step Solution

Detailed explanation

Given that \(b\) is the arithmetic mean of \(a\) and \(c\), we have \(b = \dfrac{a+c}{2}\), which implies \(2b = a+c\) or \(c = 2b - a\). The terms \((b-a), (c-b), a\) are in geometric progression. Therefore, \((c-b)^2 = (b-a) \times a\). Substituting \(c = 2b - a\) into the…