Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 815: 43


$a_n=80 (-\dfrac{1}{2})^{n-1}$

Work Step by Step

The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, we have $a_2=a_1 r \implies r=\dfrac{-40}{a_1}$ and $a_4=a_1 r^{4-1} \implies r=\dfrac{-40}{a_1}$ Equation (1) gives: $-10=a_1 \times (\dfrac{-40}{a_1})^3$ $-10=a_1 \times \dfrac{-64000}{a_1^3}$ This gives: $a_1=\sqrt {6400}=80$ Hence, $a_n=80 (-\dfrac{1}{2})^{n-1}$
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