One root common to $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ 
If three distinct numbers $a,b,c$ are in GP, and the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root, then which of the following statements is correct?  $1.$ $d,e,f$ are in GP.  $2.$ $d,e,f$ are in AP.  $3.$ $\frac da,\frac eb, \frac fc$ are in GP.  $4.$ $\frac da,\frac eb, \frac fc$ are in AP.

My attempt:
Let $r$ be the common ratio in $a,b,c$. So, $b=ar, c=ar^2$.
So, the first equation becomes $ax^2+2arx+ar^2=0\implies x^2+2rx+r^2=0$.
Let $\alpha$ be the common root. So, $\alpha^2+2r\alpha+r^2=0$. Also, $d\alpha^2+2e\alpha+f=0\implies \alpha^2+2\frac ed\alpha+\frac fd=0$.
On comparing, I get $r=\frac ed, r^2=\frac fd\implies(\frac ed)^2=\frac fd\implies e^2=fd$.
So, I am getting option $1$ as correct. But the answer is given as $4$. What's my mistake?
While this post indeed has a lot of good answers, my question was about my mistake in the method I followed. Scilife has answered that in the comments below.
 A: Following your notation, let us say that the common ratio of the geometric progression is $r$. The equation, as you currently noted, becomes:
$$ x^2 + 2rx + r^2 = 0 $$
It can easily be seen that the above equation has a double root, which is $-r$. Hence, this must be the root in common with the second quadratic equation as well.
Substituting the root into the second equation, we get:
$$dr^2 - 2er + f = 0$$
Using the fact that $r^2 = \frac{c}{a}$ and $r = \frac{c}{b}$ and dividing both sides of the equation by $c$, we see that:
$$\frac{d}{a} - 2\frac{e}{b} + \frac{f}{c} = 0 $$
Which is the condition for 3 numbers to be in arithmetic progression.
A: From the progression $ \ a \ , \ b = ar \ , \ c = ar^2 \ \ , $ we can write the first quadratic polynomial as
$$   ax^2 \ + \ 2bx \ + \ c \ \ = \ \ ax^2 \ + \ 2·ar·x \ + \ ar^2 \ \ , \ \ $$
for which we immediately see that it is a binomial-square with a "double zero" $ \ -r \ \ , $ as Scilife has observed.  For the second polynomial, we label its other zero as $ \ s \ $ and may then express it as
$$   dx^2 \ + \ 2ex \ + \ f \ \ = \ \ d·(x + r)·(x - s)  \ \ = \ \ dx^2 \ + \ d·(r - s)·x \ + \ (-d·r·s) \ \ . $$
Since the choice of $ \ s \ $ is unspecified, there is no general statement we can make about what sort of progression $ \ d \ , \ \frac12 d·(r - s) \ , \ -d·rs \ \ $ constitutes.  To investigate the other choices for this question, we might directly construct the ratios
$$ \frac{e}{b} \ \ = \ \ \frac{\frac12·d \ · \ (r \ - \ s)}{a·r} \ \ = \ \  \left(\frac12 \ - \ \frac{s}{2r} \right)·\frac{d}{a} \ \ \ , \ \ \ \frac{f}{c} \ \ = \ \ \frac{-d· \ r ·s}{a·r^2} \ \ = \ \   -   \frac{s}{r}  ·\frac{d}{a} \ \ .   $$
This form for the expressions then makes it evident that
$$ \frac{\frac{d}{a} \ + \ \frac{f}{c}}{2} \ = \ \frac12  ·  \left( \ \frac{d}{a} \ + \ \left[ \ -   \frac{s}{r}  ·\frac{d}{a} \ \right] \ \right) \ \ = \ \ \frac12  · \frac{d}{a} · \left( \ 1 \   - \ \frac{s}{r}  \ \right) \ \ = \ \ \frac{e}{b} \ \ , $$
indicating that choice $ \ \mathbf{(4)} \ $ is correct.
