Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.5 Use Recursive Rules with Sequences and Functions - 12.5 Exercises - Skill Practice - Page 831: 42b

Answer

The sequence $12,6,3$ repeats indefinitely

Work Step by Step

Consider $a_1=7$ and calculate the first ten terms: $$\begin{align*} a_1&=7\\ a_2&=3(7)+3=24\\ a_3&=\dfrac{24}{2}=12\\ a_4&=\dfrac{12}{2}=6\\ a_5&=\dfrac{6}{2}=3\\ a_6&=3(3)+3=12\\ a_7&=\dfrac{12}{2}=6\\ a_8&=\dfrac{6}{2}=3\\ a_9&=3(3)+3=12\\ a_{10}&=\dfrac{12}{2}=6. \end{align*}$$ We notice that the sequence $12,6,3$ repeats indefinitely. Consider $a_1=8$ and calculate the first ten terms: $$\begin{align*} a_1&=8\\ a_2&=\dfrac{8}{2}=4\\ a_3&=\dfrac{4}{2}=2\\ a_4&=\dfrac{2}{2}=1\\ a_5&=3(1)+3=6\\ a_6&=\dfrac{6}{2}=3\\ a_7&=3(3)+3=12\\ a_8&=\dfrac{12}{2}=6\\ a_9&=\dfrac{6}{2}=3\\ a_{10}&=3(3)+3=12. \end{align*}$$ We notice that the sequence $12,6,3$ repeats indefinitely. Consider $a_1=9$ and calculate the first ten terms: $$\begin{align*} a_1&=9\\ a_2&=3(9)+3=30\\ a_3&=\dfrac{30}{2}=15\\ a_4&=3(15)+3=48\\ a_5&=\dfrac{48}{2}=24\\ a_6&=\dfrac{24}{2}=12\\ a_7&=\dfrac{12}{2}=6\\ a_8&=\dfrac{6}{2}=3\\ a_9&=3(3)+3=12\\ a_{10}&=\dfrac{12}{2}=6. \end{align*}$$ We notice that the sequence $12,6,3$ repeats indefinitely.
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