College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises - Page 615: 58



Work Step by Step

We are asked to find the sum of: $10800+1080+108+...+0.000108$ We see that this is a geometric sequence with $a_1=10800$. We find $r$: $r=1080/10800=0.1$ We know that a geometric sequence has the form: $a_{n}=ar^{n-1}$ We use this to find $n$: $a_n=(10800)(0.1)^{n-1}$ $0.000108=(10800)(0.1)^{n-1}$ $10^{-8}=(0.1)^{n-1}$ $10^{-8}=(10)^{-(n-1)}$ $10^{-8}=(10)^{1-n}$ $-8=1-n$ $n=1+8=9$ We know the partial sum of a geometric sequence is: $S_n=a_1\frac{1-r^n}{1-r}$ $S_{9}=10800 \frac{1-(\frac{1}{10})^{9}}{1-\frac{1}{10}}=11,999.999988$
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