## College Algebra (6th Edition)

Published by Pearson

# Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.3 - Page 740: 41

#### Answer

The Sum of the given infinite geometric series = $\frac{2}{3}$

#### Work Step by Step

The Sum of a infinite geometric series ( if | r | $\lt$ 1 ) is given by S = $\frac{First term }{1 - common ratio}$ = $\frac{a_{1}}{1 - r}$ The given infinite geometric series = 1 - $\frac{1}{2}$ + $\frac{1}{4}$ - $\frac{1}{8}$ + ...................... Here First term $a_{1}$ = 1 common ratio r = $\frac{-\frac{1}{8}}{\frac{1}{4}}$ = $\frac{\frac{1}{4}}{-\frac{1}{2}}$ = $\frac{-\frac{1}{2}}{1}$ = - $\frac{1}{2}$ The Sum of the given infinite geometric series = $\frac{a_{1}}{1 - r}$ = $\frac{1}{1 - (-\frac{1}{2})}$ = $\frac{1}{1 + \frac{1}{2}}$ = $\frac{1}{\frac{3}{2}}$ = $\frac{2}{3}$

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