Chapter 11 - 11.2 - Arithmetic Sequences and Partial Sums - 11.2 Exercises - Page 786: 28

$a_n=5n-9$

The nth term of an arithmetic sequence can be obtained by the following formula: $a_n=a_1+(n-1)d$, where $a_1$ is the first term and $d$ is the common difference. $a_5=a_1+(5-1)d=a_1+4d.$ Thus $a_5-a_1=4d=16-(-4)=20\\d=5$ Hence here: $a_1=-4,d=5$ Thus $a_n=a_1+(n-1)d=-4+(n-1)5=-4+5n-5=5n-9$