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:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$

Hint:
Eliminate $c$
$$2ab=(a+b)c=(a+b)(2b-a)$$
Simplify to find $$0=a^2+ab-2b^2=(a+2b)(a-b)$$
