Chapter 9 - Conic Sections, Sequences, and Series - 9.5 Series - 9.5 Exercises - Page 747: 26

$-3198$

$a_{n+1}-a_n=-8(n+1)-15-(-8n-15)=-8$ Since the difference between consecutive terms is constant, the given sequence is arithmetic. The formula for the finite sum of the arithmetic sequence is: $S_{n}=\frac{n(a_{1}+a_{n})}{2}$ $a_{n}=-8n-15$ $a_{1}=-8(1)-15=-23$ . $a_{26}=-8(26)-15=-223$ $S_{26}=\frac{26(a_{1}+a_{26})}{2}=\frac{26(-23-223)}{2}=-3198$