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: 74

Answer

convergent; $S_{\infty}=-\dfrac{1000}{117}$

Work Step by Step

RECALL: (1) The sum of an infinite geometric series is convergent if $|r| < 1$. The sum is given by the formula: $S_{\infty}=\dfrac{a}{1-r}$ (2) An infinite geometric series is divergent if $|r|\ge1$. (3) The common ratio $r$ of a geometric series can be found by dividing any term by the term before it: $r = \dfrac{a_n}{a_{n-1}}$ Solve for $r$ to obtain: $\require{cancel} r = \dfrac{\frac{10}{3}}{-\frac{100}{9}} \\r=\dfrac{10}{3} \cdot \left(-\dfrac{9}{100}\right) \\r=\dfrac{\cancel{10}}{\cancel{3}} \cdot \left(-\dfrac{\cancel{9}3}{\cancel{100}10}\right) \\r=-\dfrac{3}{10}$ Since $|-\frac{3}{10}|<1$, then the series is convergent. Solve for the sum using the formula above, with $a=-\dfrac{100}{9}$ and $r=-\dfrac{3}{10}$, to obtain: $\require{cancel} S_{\infty}=\dfrac{-\frac{100}{9}}{1-(-\frac{3}{10})} \\S_{\infty}=\dfrac{-\frac{100}{9}}{\frac{10}{10}+\frac{3}{10}} \\S_{\infty}=\dfrac{-\frac{100}{9}}{\frac{13}{10}} \\S_{\infty} = -\dfrac{100}{9} \cdot \dfrac{10}{13} \\S_{\infty}=-\dfrac{1000}{117}$
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