Answer
n-th term:$\quad a_{n}=(-1)^{n}\cdot 4n$
The next term is $\quad a_{5}=-20$
Work Step by Step
The sign alternates, $(-,+,-,+,-,....)$
We use the fact that the signs of powers of $(-1)$ also alternate.
$\left[\begin{array}{cccccccc}
n: & 1 & 2 & 3 & 4\\
a_{n}: & (-1)^{1}\cdot 4 & (-1)^{2}\cdot 8, & (-1)^{3}\cdot 12, & (-1)^{4}\cdot 16
\end{array}\right]$
$\Rightarrow (-1)^{n}$ is a factor of the nth term
(The nth term is negative for odd n's and positive for even n's)
The absolute values of the terms are multiples of 4,
$\left[\begin{array}{cccccccc}
n: & 1 & 2 & 3 & 4 & ...\\
a_{n}: & (-1)^{1}\cdot 4\cdot 1 & (-1)^{2}\cdot 4\cdot 2, & (-1)^{3}\cdot 4\cdot 3, & (-1)^{4}\cdot 4\cdot 4 &
\end{array}\right]$
$\Rightarrow (4n)$ is a factor of the nth term
The n-th term is: $a_{n}=(-1)^{n}\cdot 4n$
The next term is $a_{5}=-4(5)=-20$
