<s>[INST] Is there any flaw in my working of showing that $-a,2b,2c$ are in Geometric Progression 
If $a,b,c$ are in Arithmetic Progression and $a^2,b^2,c^2$ are in Harmonic Progression then which of the following is true$?$

$1.$ $a=b=c$
$2.$ $2a^2=b^2=c^2$
$3.$ $a,b,c$ are in Geometric Progression
$4.$ $-a,2b,2c$ are in Geometric Progression

My work:
$$\frac{1}{b^2}-\frac{1}{a^2}=\frac{1}{c^2}-\frac{1}{b^2}$$
$$\frac{(a+b)(a-b)}{a^2b^2}=\frac{(c+b)(b-c)}{c^2b^2}$$
$$\frac{a+b}{a^2}=\frac{b+c}{c^2}$$
$$ac^2+bc^2-ba^2-ca^2=0$$
$$-(a-c)(ab+bc+ca)=0$$
$$(a-c)(ab+bc+ca)=0$$
If we consider, $a=c$ case then options $1$ and $3$ are true.
Certainly option $2$ cannot hold in any case as then the common difference of the Arithmetic Progression will not be same.
So now we drew the conclusions of three options. But here comes the twist.
If we consisder$$ab+bc+ca=0$$
$$b(a+c)+ca=0$$
$$2b^2=-ac$$
$$b^2=\frac{-a}{2}\cdot c$$
$\implies$ $\frac{-a}{2},b,c$ are in Geometric Progression
$$\implies -a,2b,2c \:\:\textrm{are in Geometric Progression} $$
Hence we conclude that optionn $4$ can also be true. But my book gives only options $1$ and $3$ as answers.
Why is this so$?$ Is there any flaw in my working$?$
Any help is greatly appreciated.
 [/INST] Hint:
Clearly $a=b=c$ is a solution
Otherwise let's try another way:
WLOG  $a=b-d, c=b+d, abcd\ne0$
$$\dfrac1{b^2}-\dfrac1{a^2}=\dfrac1{c^2}-\dfrac1{b^2} \iff c^2(a^2-b^2)=b^2(b^2-c^2)$$
$$\iff(b+d)^2((b-d)^2-b^2)=b^2(b^2-(b+d)^2)$$
$$\iff d(b+d)^2(d-2b)=-db^2(2b+d)\iff (b+d)^2(2b-d)=b^2(2b+d)$$
$$\iff\dfrac{2b-d}{2b+d}=\dfrac{b^2}{(b+d)^2}$$
Using Componendo and Dividendo,
$$ \dfrac{2b}d=\dfrac{2b^2+2bd+d^2}{d(2b+d)}$$
$$2b(2b+d)=2b^2+2bd+d^2\iff2b^2=d^2$$
$$a,c\in b(1\pm\sqrt2)$$
Now check for the options by replacing the values of $b,c$ .
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