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: 13


$-\dfrac{1}{549, 755, 813.888}$

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}$ We are given that $ a_1=1000, r=-\dfrac{1}{2}; n=40$ Now, $ a_{40}=1000 \times (-\dfrac{1}{2})^{40-1}=1000 (-\dfrac{1}{2})^{39}$ $\implies a_{40}=-\dfrac{1}{549, 755, 813.888}$
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