## Precalculus (6th Edition) Blitzer

$3,1,3,1,3$
Given $a_1=3, a_n=-a_{n-1}+4$, we have: $a_1=3$ $a_2=-a_{1}+4=-3+4=1$ $a_3=-a_{2}+4=-1+4=3$ $a_4=-a_{3}+4=-3+4=1$ $a_5=-a_{4}+4=-1+4=3$