Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 64


$\displaystyle \sum_{j=0}^{50}10(\frac{2}{3})^{n-1}\approx45$

Work Step by Step

$\displaystyle \sum_{j=0}^{50}10(\frac{2}{3})^{n-1}=10(\frac{2}{3})^{-1}+10(\frac{2}{3})^{0}+10(\frac{2}{3})^1+...+10(\frac{2}{3})^{49}$ There are 51 terms is this sequence. $a_1=10(\frac{2}{3})^{-1}=10(\frac{3}{2})=15$ $r=\frac{a_2}{a_1}=\frac{10(\frac{2}{3})^0}{10(\frac{2}{3})^{-1}}=\frac{2}{3}$ $S_n=a_1(\frac{1-r^n}{1-r})$ $S_{51}=15(\frac{1-(\frac{2}{3})^{51}}{1-(\frac{2}{3})})=15(\frac{1-(\frac{2}{3})^{51}}{\frac{1}{3}})\approx45$
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.