The first four terms x, y, z, w of an arithmetic sequence I have been attempting this question for the past 3 days with no luck:
The first four terms x, y, z, w of an arithmetic sequence satisfy x + y + z + w = 8 and xw + yz = -2. Find all possible values of x, y, z, w.
My attempt:
I know the key here is finding the difference between the terms but I am having a hard time figuring out how to find this. I tried:
y - x = d
z - y = d 
w - z = d
I then substitue the second equation in to the first equation
y - x = z - y
I am actually not sure if this is the right method though because also I have thought about adding or subtracting all the equations together but I was not able to get a significant solution out of it. I just need a little guidance on how to find d the difference. Thanks for your time.
 A: I would look for what symmetry can do for me.  Let $d$ be the common difference, and let $m$ be the "midpoint," that is, the point halfway between $y$ and $z$. Let $d=2e$ (it is nice to avoid fractions).
Then $y=m-e$, $z=m+e$, $x=m-3e$, and $w=m+3e$.  The sum is $4m$, so $m=2$.
The second equation says that $(m^2-9e^2)+(m^2-e^2)=-2$.  Thus $2m^2-10e^2=-2$, and therefore $e^2=1$.  It follows that $e=\pm 1$, and now we know everything. Two arithmetic sequences satisfy the equations,  $(-1, 1, 3, 5)$ and its reverse $(5,3,1,-1)$. 
Comment: We can let the first term be $a$, and the common difference $d$ (these are the usual names). Then we have $x=a$, $y=a+d$, $z=a+2d$, and $w=a+3d$.  Substitute in our two equations. We get a system of two equations in the two unknowns $a$ and $d$, one linear and the other quadratic.   The system is not difficult to solve, but the calculation is not as immediate as the symmetrical one based on $m$ and $e$.  Symmetry is our friend!
A: i guess there  would be a list of solutions... take the first term and the other consecutive would be written in term,s of the first term and the common difference... then on so;ving we get .. 37d^2  +(3a-96)d  +  65=0....
