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 912: 28

Answer

$\$1710.34$

Work Step by Step

The provided sequence is $\$1000,\$1050,\$1102.50,\ldots$, Therefore, Find the common ratio $r$, by using the formula $r=\frac{{{a}_{2}}}{{{a}_{1}}}$. $\begin{align} & r=\frac{1050}{1000} \\ & =1.05 \end{align}$ The general formula is ${{a}_{n}}=a{{r}^{n-1}}$. Substitute the value of the first term and the common ratio in the equation ${{a}_{n}}=a{{r}^{n-1}}$, where $n=12$. $\begin{align} & {{a}_{12\text{th}}}=1000{{\left( 1.05 \right)}^{12-1}} \\ & =1000{{\left( 1.05 \right)}^{11}} \end{align}$ So, the value obtained is $1710.34$ Thus, the value of the $12\text{th}$ term of the geometric sequence $\$1000,\$1050,\$1102.50,\ldots$ is $\$1710.34$.
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