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: 61


$\displaystyle \sum_{j=0}^{5}200(1.05)^n=1360.38$

Work Step by Step

$\displaystyle \sum_{j=0}^{5}200(1.05)^n=200(1.05)^0+200(1.05)^1+200(1.05)^2+...200(1.05)^5$ There are 6 terms is this sequence. $a_1=200(1.05)^0=200$ $r=\frac{a_2}{a_1}=\frac{200(1.05)^1}{200(1.05)^0}=1.05$ $S_n=a_1(\frac{1-r^n}{1-r})$ $S_{6}=200(\frac{1-(1.05)^{6}}{1-1.05})=200(\frac{1.05^6-1}{0.05})=1360.38$
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