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Suppose $a, b, c$ are distinct positive real numbers such that $a, 2 b, 3 c$ are in A.P. and $a, b, c$ are in G.P. The common ratio of G.P. is

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2

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3

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detailed solution

Correct option is C

We have $2 (2 b) = a + 3 c$ and $b^{2} = a c$
$\begin{array}{l} ∴ \frac{1}{16} (a + 3 c)^{2} = a c \\ \Rightarrow a^{2} + 9 c^{2} + 6 a c = 16 a c \\ \Rightarrow a^{2} - 10 a c + 9 c^{2} = 0 \\ \Rightarrow (a - c) (a - 9 c) = 0 \Rightarrow a = 9 c [∵ a \neq c] \end{array}$
$\Rightarrow \frac{c}{a} = \frac{1}{9}$ or $r^{2} = \frac{1}{9} \Rightarrow r = \frac{1}{3}$

$[∵ a, b, c > 0]$

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