## Algebra 2 (1st Edition)

$\left\{\begin{array}{l} a_{1}=15,\\ a_{n}=a_{n-1}-4 \end{array}\right.$
$a_{1}=15$ 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: $\left\{\begin{array}{l} a_{1}=15,\\ a_{n}=a_{n-1}-4 \end{array}\right.$