Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 10



Work Step by Step

Need to find the first four terms of $a_n=\dfrac{3n}{n+5}$ when $n=1,2,3,4$ Thus, $a_1=\dfrac{3(1)}{1+5}=\dfrac{3}{6}=\dfrac{1}{2}$ $a_2=\dfrac{3(2)}{2+5}=\dfrac{6}{7}$ $a_3=\dfrac{3(3)}{3+5}=\dfrac{9}{8}$ $a_4=\dfrac{3(4)}{4+5}=\dfrac{12}{9}$ Hence, the first four terms are: $\dfrac{1}{2},\dfrac{6}{7},\dfrac{9}{8},\dfrac{12}{9}$
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