If $a,b,c$ are in Geometric Progression and$ a-b,c-a,b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to? If $a,b,c$ are in Geometric Progression and $a-b$,$c-a$,$b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to?
The statements I could get are
1)$b^2=ac$
2)$(c-a)^2=2(b-a)(b-c)$
I get that $(a+4b+c)^2=24ac+6ab+6bc$.
how to proceed and get to the answer??Options given are 
a)1
b)0
3)2abc
d)b^2+ac

The condition for a harmonic progression is that
$$ \frac{1}{a-b} + \frac{1}{b-c} = \frac{ 2}{ c-a}. $$
Clearing denominators, we get
$$ a^2 -2b^2 + c^2 + 2ab + 2 bc - 4ac = 0. $$
This gives us 
$$(a+b+c)^2 = a^2 + b^2 +c^2 + 2ab + 2bc + 2ac = 0 + 3b^2 + 6ac =9b^2.$$
Hence, either $ a+b+c = 3b$ or $a+b+c = -3b$.
Now, in the first case, since $a+c = 2b$ and $ac = b^2$, hence we must have $a=b=c$. This means that $a-b, c-a, b-c$ is not a harmonic progression.
Thus, we must have $a+b+c = - 3b$, or that $a+4b+c = 0$.
