Finding the next term in a sequence. So this came up in a math competition some of students participated in over the weekend.  Expected highest level of mathematics is calculus 1.  
Let $a_n$ be an arithmetic sequence, $b_n$ be a geometric sequence and suppose that $c_n=a_n-b_n$.  The first four terms of $c_n$ are $2, 8, 6, 20$.  Find $c_5$.
My students had two minutes and no calculator to answer this... I've played with this a bit and think that partial sums is probably the direction to go if this is going to be a 2 minute problem but I'm stuck!  I've gotten to a couple spots that I think are along the right direction but perhaps I'm not thinking about this correctly.  Here's what I know:
$c_5=S_5-S_3-c_4$
So if I could find the partial sum at 5 I've got this... The other idea that I have is that the common ratio of the geometric series is $(-b_2)/(b_1)$ (I believe) but I'm still fairly at a loss as how to get our next term.  Some help would be appreciated!
 A: Let $d$ denote the common difference of the arithmetic sequence and $r$ denote the common ratio of the geometric sequence.  Then:
\begin{align*}
  c_1 &= a_1 - b_1\\
  c_2 &= a_2 - b_2\\
    &= (a_1 + d) - (b_1r)\\
  c_3 &= a_3 - b_3\\
    &= (a_1 + 2d) - (b_1r^2)\\
  c_4 &= a_4 - b_4\\
    &= (a_1 + 3d) - (b_1r^3)
\end{align*}
And you want to find this value:  $c_5 = a_5-b_5 = (a_1+4d) - (b_1r^4)$.
We have this system:
\begin{align*}
  2 &= a_1 - b_1\\
  8 &= a_1 + d - b_1r\\
  6 &= a_1 + 2d - b_1r^2\\
  20 &= a_1 + 3d - b_1r^3
\end{align*}
It's mildly unpleasant but it's four equations with four variables, and there is indeed a unique solution.  Can you take it from here?
A: I don't think my solution can be calculated within two minutes, but it is pretty straightforward and can be calculated fast too. Let $a_i = x+(i-1)\cdot y$. Since $b_i = a_i-c_i$, we have $$b_2^2 = b_1b_3\implies (x+y-8)^2=(x-2)\cdot(x+2y-6)\implies y^2-12y+52=8x$$
$$b_3^2=b_2b_4 \implies (x+2y-6)^2=(x+y-8)(x+3y-20)\implies y^2+20y-124=-16x$$
Multiplying the first equation by $2$ and adding the second equation, $$3y^2-4y-20=0\implies y=-2,x=10\ \ \text{or} \ \ \ y=\frac{10}{3}, x=\frac{26}{9}$$
The first case is not what we are looking for because it gives $b_2=b_3=0$, so the second case must be true.
Hence, we can get $a_5=x+4y=\frac{146}{9}$. Further, $$b_1=a_1-c_1=\frac{8}{9}, b_2 = a_2-c_2 = -\frac{16}{9}\implies b_5 = \frac{128}{9}$$ Thus, $$c_5=a_5-b_5=\frac{18}{9}=2$$
