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If $a, b, c$ are in AP, $b-a, c-b$ and $a$ are in GP, then $a: b: c$ is
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The correct answer is:
$1: 2: 3$
Given, that, $2 b=a+c$ and
$(c-b)^2=(b-a) a$
$\begin{aligned} & \therefore \quad(b-a)^2=(b-a) a \\ & \Rightarrow \quad b=2 a \\ & \Rightarrow \quad c=3 a \\ & \Rightarrow \quad a: b: c=1: 2: 3 \\ & \end{aligned}$
$(c-b)^2=(b-a) a$
$\begin{aligned} & \therefore \quad(b-a)^2=(b-a) a \\ & \Rightarrow \quad b=2 a \\ & \Rightarrow \quad c=3 a \\ & \Rightarrow \quad a: b: c=1: 2: 3 \\ & \end{aligned}$
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