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 - 14.1 Sequences and Series - 14.1 Exercise Set - Page 895: 89

Answer

$\$2500,\$2000,\$1600,\$1280,\$1024,\$819.20,\$655.36,\$524.29,\$419.43,\$335.54$.

Work Step by Step

Consider the value of an LCD projector, $\$2500$. Its resale value decreases by 20% each year. Thus, the value for the second year is 80% of the value for the first year. For the second year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 2500 \\ & =80\cdot 25 \\ & =2000 \end{align}$ For the third year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 2000 \\ & =80\cdot 20 \\ & =1600 \end{align}$ For the fourth year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 1600 \\ & =80\cdot 16 \\ & =1280 \end{align}$ For the fifth year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 1280 \\ & =80\cdot 128 \\ & =1024 \end{align}$ For the sixth year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 1024 \\ & =80\cdot 102.4 \\ & =819.20 \end{align}$ For the seventh year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 819.20 \\ & =80\cdot 8.1920 \\ & =655.36 \end{align}$ For the 8th year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 655.36 \\ & =80\cdot 6.5536 \\ & =524.29 \end{align}$ For the 9th year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 524.29 \\ & =80\cdot 5.2429 \\ & =419.43 \end{align}$ For the tenth year: $\begin{align} & \text{LCD}=\frac{80}{100}\cdot 419.43 \\ & =80\cdot 4.1943 \\ & =355.54 \end{align}$ Thus, the sequence for the resale value is: $\$2500,\$2000,\$1600,\$1280,\$1024,\$819.20,\$655.36,\$524.29,\$419.43,\$335.54$.
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