If $a,b,c$ be in Arithmetic Progression, If $a,b,c$ be in Arithmetic Progression, $b,c,a$ in Harmonic Progression, prove that $c,a,b$ are in Geometric Progression.
My Attempt:
$a,b,c$ are in AP so
$$b=\dfrac {a+c}{2}$$
$b,c,a$ are in HP so
$$c=\dfrac {2ab}{a+b}$$
Multiplying these relations:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$

Since $a,b,c$ are in arithmetic progression, we get
\begin{align*}
&c-b=b-a\\[4pt]
\implies\;&a = 2b - c\\[4pt]
\end{align*}
Since $b,c,a$ are in harmonic progression, we get
\begin{align*}
&\frac{1}{a}-\frac{1}{c} = \frac{1}{c}-\frac{1}{b}\\[4pt]
\implies\;&a = \frac{bc}{2b-c}\\[4pt]
\implies\;&a = \frac{bc}{a}\\[4pt]
\implies\;&\frac{a}{c}=\frac{b}{a}\\[4pt]
\end{align*}
hence, $c,a,b$ are in geometric progression, as was to be shown.
