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: 34


$r = \dfrac{1}{4}$ $a_5=-\dfrac{1}{32}$ The $n^{th}$ term of the geometric sequence is: $a_n=-8 \cdot \left(\dfrac{1}{4}\right)^{n-1}$

Work Step by Step

RECALL: (1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms: $r =\dfrac{a_n}{a_{n-1}}$ (2) The $n^{th}$ term of a geometric sequence is given by the formula: $a_n = a\cdot r^{n-1}$ where $a$ = first term $r$ = common ratio The sequence is said to be geometric. Thus, we can proceed to solving for the common ratio: $r=\dfrac{-2}{-8} \\r = \dfrac{1}{4}$ The fifth term can be found by multiplying the common ratio to the fourth term. The fourth term is $-\dfrac{1}{8}$. Thus, the fifth term is: $a_5=-\dfrac{1}{8} \cdot \left(\dfrac{1}{4}\right) \\a_5=-\dfrac{1}{32}$ With a first term of $-8$ and a common ratio of $r=\dfrac{1}{4}$, the $n^{th}$ term of the geometric sequence is: $a_n=a \cdot r^{n-1} \\a_n=-8 \cdot \left(\dfrac{1}{4}\right)^{n-1}$
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