Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1073: 22

Answer

$ a_{n}=5(\dfrac{-1}{5})^{n-1}$ and $ a_{7}=\dfrac{1}{3125}$

Work Step by Step

The general formula to find the nth term of a Geometric sequence is given as: $ a_{n}=a_1r^{n-1}$ $ r=\dfrac{a_2}{a_1}=\dfrac{-1}{5}$ and $ n=7$ Now, $ a_{n}=5(\dfrac{-1}{5})^{n-1}$ Now, $ a_{7}=5(\dfrac{-1}{5})^{7-1}=5(\dfrac{-1}{5})^{6}=\dfrac{1}{3125}$ Our answers are: $ a_{n}=5(\dfrac{-1}{5})^{n-1}$ and $ a_{7}=\dfrac{1}{3125}$
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