<s>[INST] 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:
$$bc=\dfrac {a+c}{2} \dfrac {2ab}{a+b}$$
$$=\dfrac {2a^2b+2abc}{2(a+b)}$$
$$=\dfrac {2a^2b+2abc}{2a+2b}$$
 [/INST] Since $a,b,c$ are in AP we have $a=b-x$ and $c=b+x$ for some $x$ and since $b,c,a$ are in HP  we have $$(b+x)(2b-x) = 2(b-x)b$$
Solwing this we get: $x=0$ or $x=3b$. So in the later case we get $a=-2b$ and $c=4b$ and so $$a^2=4b^2= bc$$
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