## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 34

#### Answer

$a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$

#### Work Step by Step

We are given: $3, 0.3, 0.03, 0.003$ ... We see that each term is the previous term divided by $10$, starting with $3$. We find the pattern: $a_1=3*10^{0}$ $a_2=3*10^{-1}$ $a_3=3*10^{-2}$ $a_4=3*10^{-3}$ Therefore: $a_{n}=3*(\frac{1}{10})^{n-1}$ Or: $a_{n}=3*(10)^{1-n}$

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