Answer
$\left\{\begin{array}{l}
a_{1}=15,\\
a_{n}=a_{n-1}-4
\end{array}\right.$
Work Step by Step
$a_{1}=15$
We obtain $a_{2}=11$ from $a_{1}=15$ by adding $(-4)$ to it.$ \quad a_{2}=a_{1}-4$
We obtain $a_{3}=7$ from $a_{2}=11$ by adding $(-4)$ to it.$ \quad a_{3}=a_{2}-4$
We obtain $a_{4}=3$ from $a_{3}=7$ by adding $(-4)$ to it.$ \quad a_{4}=a_{3}-4$
We obtain $a_{5}=-1$ from $a_{4}=3$ by adding $(-4)$ to it.$ \quad a_{5}=a_{4}-4$
We obtain $a_{6}=-5$ from $a_{4}=3-1$ by adding $(-4)$ to it.$ \quad a_{5}=a_{4}-4$
$...$
We obtain $a_{n}$ from $a_{n-1}$ by adding $(-4)$ to it.$ \quad a_{n}=a_{n-1}-4$
A recursive rule for the sequence:
$\left\{\begin{array}{l}
a_{1}=15,\\
a_{n}=a_{n-1}-4
\end{array}\right.$
