Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1073: 4

Answer

$24,8, \dfrac{8}{3},\dfrac{8}{9}, \dfrac{8}{27}$

Work Step by Step

The general formula to find the nth term of a Geometric sequence is given as: $ a_{n+1}=a_nr $ We are given that $ a_1=24$ $ a_2=a_1r = (24)(\dfrac{1}{3})=8 ;\\ a_3=a_2r = (8)(\dfrac{1}{3})=\dfrac{8}{3}; \\ a_4=a_3r = (\dfrac{8}{3})(\dfrac{1}{2})=\dfrac{8}{9} ; \\ a_5=a_4r = (\dfrac{8}{9})(\dfrac{1}{3})=\dfrac{8}{27}$ Hence, the first five terms are: $24,8, \dfrac{8}{3},\dfrac{8}{9}, \dfrac{8}{27}$
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.