<s>[INST] 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
 [/INST] Let $b=ra$ and $c=r^2a$.  Then $c-a=\frac{2(a-b)(b-c)}{a-c}$, so $(c-a)^2=2(b-a)(b-c)$ and therefore $(r^2-1)^2 a^2=2(r-1)ar(1-r)a=-2r(r-1)^2a^2$.     Dividing by
$(r-1)^2a^2$ gives $(r+1)^2=-2r$ and therefore $r^2+4r+1=0$.  Thus $a+4b+c=a+4ra+r^2 a=a(r^2+4r+1)=0$.
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