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: 25


The first five terms are: $a_1 = \dfrac{1}{4} \\a_2 = \dfrac{1}{16} \\a_3 = \dfrac{1}{64} \\a_4=\dfrac{1}{256} \\a_5=\dfrac{1}{1024}$ The sequence is geometric with a common ratio of $\dfrac{1}{4}$. The $n^{th}$ term is :$a_n = \dfrac{1}{4} \cdot\left(\dfrac{1}{4}\right)^{n-1}$

Work Step by Step

To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain: $a_1 = \dfrac{1}{4^1} = \dfrac{1}{4} \\a_2 = \dfrac{1}{4^2} = \dfrac{1}{16} \\a_3 = \dfrac{1}{4^3} = \dfrac{1}{64} \\a_4=\dfrac{1}{4^4} = \dfrac{1}{256} \\a_5=\dfrac{1}{4^5} = \dfrac{1}{1024}$ RECALL: A sequence is geometric if there is a common ratio among consecutive terms. Notice that $\dfrac{1}{4}$ times a term is equal to the value of the next term. This means that the sequence has a common ratio of $\frac{1}{4}$. Thus, the sequence is geometric with a common ratio of $\dfrac{1}{4}$. The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$ where $a$ = first term and $r$= common ratio. This means that the $n^{th}$ term of the given geometric sequence, whose $a=\dfrac{1}{4}$ and $r=\dfrac{1}{4}$, is: $a_n = \dfrac{1}{4} \cdot\left(\dfrac{1}{4}\right)^{n-1}$
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