Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.2 - Arithmetic Sequences - Exercise Set - Page 1059: 8



Work Step by Step

Formula to calculate the nth term of an arithmetic sequence is: $a_n=a_1+d(n-1)$ Here, $a_1: $ First term and $d=$ common difference $a_1=\dfrac{3}{4}+(-\dfrac{1}{4})(1-1)=\dfrac{1}{2}; \\a_2=\dfrac{3}{4}+(-\dfrac{1}{4})(2-1)=\dfrac{1}{4};\\a_3=\dfrac{3}{4}+(-\dfrac{1}{4})(3-1)=0;\\a_4=\dfrac{3}{4}+(-\dfrac{1}{4})(4-1)=\dfrac{-1}{4}; \\a_5=\dfrac{3}{4}+(-\dfrac{1}{4})(5-1)=\dfrac{-1}{2};\\a_6=\dfrac{3}{4}+(-\dfrac{1}{4})(6-1)=\dfrac{-3}{4}$ Hence, the first six terms are: $\dfrac{1}{2},\dfrac{1}{4},0,\dfrac{-1}{4},\dfrac{-1}{2},\dfrac{-3}{4}$
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