Discussion
If \(a, b, c\) are three unequal numbers such that \(a, b, c\) are in arithmetic progression and \(b - a, c - b, a\) are in geometric progression, then \(a : b : c\) is
Answer: B
The ratio \(a : b : c\) is 1:2:3.
Since \(a, b, c\) are in AP, \(b - a = c - b = d\), with \(d \ne 0\) because the numbers are unequal. The terms \(b - a, c - b, a\) are then \(d, d, a\), and for these to be in GP:
\[ \begin{aligned} d^{2} &= d \cdot a \\ d &= a \end{aligned} \]
So \(b = a + a = 2a\) and \(c = 3a\), giving \(a : b : c = 1 : 2 : 3\). Check: 1, 2, 3 is an AP, and \(1, 1, 1\) is a GP. The other ratios are not even arithmetic progressions, except 3:4:5, for which \(1, 1, 3\) is not a GP.
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