Chapter 9 - Conic Sections, Sequences, and Series - 9.5 Series - 9.5 Exercises - Page 747: 23

$825$

$a_{n+1}-a_n=20+(n+1-1)5-(20+(n-1)5)=5$ As the difference between consecutive terms is constant, the given sequence is arithmetic. The formula for the finite sum of the arithmetic sequence is: $S_{n}=\frac{n(a_{1}+a_{n})}{2}$ $a_{n}=20+(n-1)5$ $a_{1}=20+(1-1)5=20$ . $a_{15}=20+(15-1)5=90$ $S_{15}=\frac{15(a_{1}+a_{15})}{2}=\frac{15(20+90)}{2}=825$