## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1075: 74

For $5$ years, the company B salary is $\$780$more than company A. #### Work Step by Step From the given information, we can observe that it is a geometric series, For company A:$30000,1\cdot 06\left( 30000 \right),1\cdot {{06}^{2}}\left( 30000 \right),\cdots $Here${{a}_{1}}=30000$and common ratio$r=1\cdot 06$and$n=5$. We use the formula for the sum of a geometric series with finite terms${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$. So;$\begin{align} & {{S}_{n}}=\frac{30000\left( 1-1\cdot {{06}^{5}} \right)}{\left( 1-1\cdot 06 \right)} \\ & {{S}_{n}}=\frac{30000\left( 1-1\cdot 33825578 \right)}{-\left( 0\cdot 06 \right)} \\ & {{S}_{n}}=\frac{30000\left( -0\cdot 33825578 \right)}{-\left( 0\cdot 06 \right)} \\ & {{S}_{n}}=169,112.77 \\ \end{align}$For company A, the total salary for 5 years is$169112.77$. For company B:$32000,1\cdot 03\left( 32000 \right),1\cdot {{03}^{2}}\left( 32000 \right),\cdots $Here${{a}_{1}}=32000$and common ratio$r=1\cdot 03$and$n=5$. We use the formula for the sum of a geometric series with finite terms${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$. So;$\begin{align} & {{S}_{n}}=\frac{32000\left( 1-1\cdot {{03}^{5}} \right)}{\left( 1-1\cdot 03 \right)} \\ & {{S}_{n}}=\frac{32000\left( 1-1\cdot 1592 \right)}{-\left( 0\cdot 03 \right)} \\ & {{S}_{n}}=\frac{32000\left( -0\cdot 1592 \right)}{-\left( 0\cdot 03 \right)} \\ & {{S}_{n}}=169892.35 \\ \end{align}$For company B, the total salary for 5 years is$169892.35$. The difference of total salary for both companies is:$169112-169892=-780\$

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