College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 57

Answer

series converges; $S_{\infty}=\dfrac{8}{5}$

Work Step by Step

RECALL: (1) The common ratio $r$ of a geometric sequence is equal to the quotient of any term and the term before it. $r=\dfrac{a_n}{a_{n-1}}$ (2) A geometric series converges if $|r| \lt 1$. The sum of a convergent infinite geometric series is given by the formula: $S_{\infty}=\dfrac{a_1}{1-r}$ where $r$ = common ratio $a_1$ = first term $\bf\text{Solve for r}:$ $r=\dfrac{a_2}{a_1} = \dfrac{-\frac{1}{2}}{2}=-\dfrac{1}{2} \cdot \dfrac{1}{2}=-\dfrac{1}{4}$ Since $|-\frac{1}{4}|=\frac{1}{4} \lt 1$, the infinite geometric series converges. $\bf\text{Find the sum of the infinite geometric series}:$ With $a_1 = 6$ and $r=-\frac{1}{4}$, $S_{\infty} = \dfrac{a_1}{1-r} \\S_{\infty}= \dfrac{2}{1-(-\frac{1}{4})} \\S_{\infty}=\dfrac{2}{\frac{4}{4} + \frac{1}{4}} \\S_{\infty}=\dfrac{2}{\frac{5}{4}} \\S_{\infty}= 2 \cdot \dfrac{4}{5} \\S_{\infty}= \dfrac{8}{5}$
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