Chapter 9 - Conic Sections, Sequences, and Series - 9.5 Series - 9.5 Exercises - Page 746: 4

$9+6+3+0+(-3)=15$

We need to sum the first five terms of the sequence whose general term is $a_{n}=-3n+12$. This yields $\sum\limits^{5}_{n=1}(-3n+12)=[-3(1)+12]+[-3(2)+12]+[-3(3)+12]+[-3(4)+12]+[-3(5)+12]$ $=9+6+3+0+(-3)=15$