Find the expressions for the common difference and common ratio 
The first 2 terms of a geometric progression (first term $a$ and common ratio $r$) are the same as the first 2 terms of an arithmetic progression (first term $a$ and common difference $d$). The third term of the geometric progression is twice as big as the third term of the arithmetic progression. Find 2 different expressions for $d$ and hence or otherwise find the 2 possible values of $r$.

So I've tried to tackle this problem, to no avail:
$$ ar = a + d$$
$$ar^2 = 2(a + 2d)$$
So using this I found $d = ar -a $ and $d = \dfrac{1}{4}ar^2 - \dfrac{1}{2}a$, so $$ ar -a =\dfrac{1}{4}ar^2 - \dfrac{1}{2}a$$ 
$$a(r^2-4r+2)=0$$
Here I knew I was wrong, because I'm sure that you're not supposed to be using the quadratic for this exercise (judging from how much time you have).
So where did I go wrong?

I'm afraid that you've done nothing wrong. (There's an odd thing to say.)
It's worth noting that if $a=0,$ then we get $d=0$ immediately, but $r$ can take literally any value. Since the problem expects two possible values for $r,$ then we clearly must assume $a\neq0.$
Since $a\ne 0,$ then $r^2-4r+2=0,$ and from there, the quadratic formula will give you the two possible values for $r$. Alternately, you could note the following: $$r^2-4r+2=0\\r(r-4)+2=0\\(r-2+2)(r-2-2)+2=0\\(r-2)^2-2^2+2=0\\(r-2)^2-2=0$$
This is an end-around method of completing the square by using the difference of squares formula, from which you can readily conclude that $r=2\pm\sqrt2,$ if you'd rather avoid the quadratic formula.
