Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1075: 80

Answer

A) $\$171271$. B) $\$135271$.

Work Step by Step

(a) In the given case $P=\$50$ and $r=0.065$ for $t=40\,\text{years}$. It is compounded monthly, hence $n=12$. Thus: $\begin{align} & A=\frac{75\left[ {{\left( 1+\frac{0.065}{12} \right)}^{12\times 40}}-1 \right]}{\left( \frac{0.065}{12} \right)} \\ & =\frac{75\left[ {{\left( 1.0054 \right)}^{12\times 40}}-1 \right]}{\left( 0.0054 \right)} \\ & =\frac{75\left[ \left( 13.3696 \right)-1 \right]}{\left( 0.0054 \right)} \\ & =171271 \end{align}$ Hence, the total savings after $65\ years$ is $\$171271$.
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