Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - Chapter Review Exercises - Page 757: 57

Answer

The sequence is geometric. $a_{n}=3\cdot 4^{n-1}$

Work Step by Step

$\dfrac{12}{3}=\dfrac{48}{12}=\dfrac{192}{48}=\dfrac{768}{192}=\dfrac{3072}{768}=4$ As $\frac{a_{n+1}}{a_{n}}=4$, that is, the ratio between each term and the preceding term is the same, the sequence is geometric. Common ratio: $r=4$. First term: $a_{1}=3$. The general term is: $a_{n}=a_{1}\cdot r^{n-1}=3\cdot 4^{n-1}$
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