Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.2 - Arithmetic Sequences - 12.2 Exercises - Page 857: 58



Work Step by Step

See p. 855. For the arithmetic sequence $a_{n}=a+(n-1)d$ the nth partial sum $S_{n}=\displaystyle \sum_{k=1}^{n}[a+(k-1)d]$ is given by either of the following equivalent formulas: 1. $S_{n}=\displaystyle \frac{n}{2}[2a+(n-1)d]\qquad $2. $S_{n}=n(\displaystyle \frac{a+a_{n}}{2})$ ------------------- We see from the given sequence that $a=-3,$ $d=\displaystyle \frac{3}{2}-0=\frac{3}{2}$ The last term is $a_{n}=30$ (we find n): $a_{n}=a+(n-1)d$ $30=-3+\displaystyle \frac{3}{2}(n-1)$ $33=\displaystyle \frac{3}{2}(n-1)\qquad/\times\frac{2}{3}$ $22=n-1$ $23=n$ And, we find $S_{23}$, (using formula 2) $S_{23}=23(\displaystyle \frac{-3+30}{2})=\frac{621}{2}=310.5$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.