Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.3 Geometric Sequences and Series - 14.3 Exercise Set - Page 913: 63

Answer

$155,797$

Work Step by Step

1. Familiarize yourself with the problem. In one year, the population will be, $100,000+0.03\left( 100,000 \right)=\left( 1.03 \right)100,000$. In two years, the population will be, $\left( 1.03 \right)100,000+\left( 0.03 \right)\left( 1.03 \right)\left( 100,000 \right)={{\left( 1.03 \right)}^{2}}100,000$ The population forms a geometry series: $100,000,\left( 1.03 \right)100,000,{{\left( 1.03 \right)}^{2}}100,000,\ldots $ Thus, the population at 15 years will be the 16th term of the geometry series. 2. Translate the problem into an equation, $\begin{align} & {{a}_{1}}=100,000 \\ & r=\frac{\left( 1.03 \right)100,000}{100,000} \\ & =1.03 \\ & n=16 \end{align}$ $\begin{align} & {{a}_{n}}={{a}_{1}}{{r}^{n-1}} \\ & {{a}_{16}}=\left( 100,000 \right){{\left( 1.03 \right)}^{16-1}} \end{align}$ 3. Carry out the mathematical operations to solve the equation. Substitute $n=16,{{a}_{1}}=100,000$ and $r=1.03$, $\begin{align} & {{a}_{n}}={{a}_{1}}{{r}^{n-1}} \\ & =\left( 100,000 \right){{\left( 1.03 \right)}^{16-1}} \\ & =\left( 100,000 \right){{\left( 1.03 \right)}^{15}} \\ & =155,797 \end{align}$ Thus, the approximate population in 15 years is $155,797$.
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