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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Aicharya · Accepted Answer

Given, that, $2 b=a+c$ and $(c-b)^2=(b-a) a$ $\begin{aligned} & 	herefore \quad(b-a)^2=(b-a) a \ & \Rightarrow \quad b=2 a \ & \Rightarrow \quad c=3 a \ & \Rightarrow \quad a: b: c=1: 2: 3 \ & \end{aligned}$

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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