Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is? 
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of
$x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP
then $p + q$ is ?

My solution approach :-
I have assumed $α,β,γ,δ$  to be as follows ;
$α = \frac{a}{r^3}$
$β = \frac{a}{r}$
$γ = ar$
$δ = ar^3$
Here the common ratio is $r^2$.
Now $α+β = 1$ and $γ+δ = 4$ and solving these two equations in terms of $a$ and $r$ gives the value of $r= \sqrt{2} \text{ and } \sqrt{-2}$ which leads to the values of $a$ to be $\frac{2\sqrt{2}}{3}$ and $\frac{-2\sqrt{2}}{3}$ respectively.
In both the above cases, $α,β,γ,δ$ turns out to be the same i.e. $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ which gives the value of $p+q = \frac{34}{9}$.
Now if I assume $α,β,γ,δ$  to be as follows ;
$α = a$
$β = ar$
$γ = ar^2$
$δ = ar^3$
Here the common ratio is just $r$.
But in this case $r$ values turn out to be $\pm2$ and for $r=2$ we get $α,β,γ,δ$ to be $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ whereas for $r=-2$ we get $α,β,γ,δ$ to be $-1,2,-4,8$ and hence $p+q = \frac{34}{9} \text{ or }-34 $.
Now I am getting confused why there is this difference in both the approaches. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!

In the first approach: when you solve for $r$ you would have got an equation $r^4=4$. This gives you $r^2=2$ and $r^2=-2$. You have only considered the first case (i.e. when $r^2=2$). If you consider the second case $r^2=-2$, then you get $r=\pm i\sqrt{2}$. In which case you will get $p+q=-34$. So both approaches will yield the same answers.

Some details:
$$\alpha+\beta=1 \implies a\frac{(1+r^2)}{r^3}=1.$$
Likewise
$$\gamma+\delta=4 \implies ar(1+r^2)=4.$$
Thus
$$r^4=4.$$
This means $r^2=\pm 2$. Consequently $a=\frac{4}{3r}$ or $a=\frac{-4}{r}$ (the second case occurs when $r^2=-2$). Now we can use,
$$\alpha \cdot \beta=p \implies p=\frac{a^2}{r^4}=\frac{a^2}{4}.$$
Likewise we have
$$q=a^2r^4=4a^2.$$
So,
$$p+q=\frac{17}{4}a^2$$
Now plug in the two possible values of $a^2=\frac{16}{9r^2}=\frac{8}{9}$ OR $a^2=\frac{16}{r^2}=-8$ to get both the possibilities.

Additional remark:
In fact, from the conditions of the problem that $p$ and $q$ are integers, we can see that $r^2=2$ is NOT a valid solution because it gives $a^2=\frac{8}{9}$, in which case $p=\frac{a^2}{4}=\frac{2}{9} \not\in \Bbb{Z}$. Thus $\color{red}{r^2=-2}$ is the ONLY valid possibility. In which case
$$\color{magenta}{p+q=-34}$$
is the ONLY valid answer.
