Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 12 - Section 12.3 - Geometric Sequences - 12.3 Exercises - Page 865: 80

Answer

$\displaystyle \frac{10,457}{4950}$

Work Step by Step

$2.11\overline{25}=2.11252525 \ldots = 2.11+ 0.00252525$ $=2.11+ 0.00\overline{25}$ $ 0.00\displaystyle \overline{25}=(\frac{25}{10,000}+\frac{25}{1,000,000}+\frac{25}{100,000,000}+\cdots )$ ... an infinite geometric series. $a=\displaystyle \frac{25}{10,000}$ , $r=\displaystyle \frac{1}{100}$. $|\displaystyle \frac{1}{100}| < 1$, so the sum exists, $S=\displaystyle \frac{a}{1-r}$ $0.00\displaystyle \overline{25}=\frac{\frac{25}{10,000}}{1-\frac{1}{100}}=\frac{25}{9900}$, so $2.11\overline{25}= 2.11+0.00\overline{25}=$ $=\displaystyle \frac{211}{100}+\frac{25}{9900}=\frac{211\cdot 99+25}{9900}=\frac{20,914}{9900}$ $=\displaystyle \frac{10,457}{4950}$
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