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 11.3 - Geometric Sequences and Series - Exercise Set - Page 855: 71


$a_2=12$ and $a_3=18$

Work Step by Step

Given: Geometric sequence: $8,a_2,a_3,27$ Here $a_4=a_1r^{n-1}=a_1r^3$ Thus, $27=8r^3 \implies r=\dfrac{3}{2}$ Now, $a_2=a_1r^{2-1}=a_1r \implies (8)(\dfrac{3}{2})=12$ and $a_3=a_1r^{3-1}=a_1r^2 \implies (12)(\dfrac{3}{2})=18$
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