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: 7



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 $a_1=\dfrac{5}{2}+(-\dfrac{1}{2})(1-1)=\dfrac{5}{2}; \\a_2=\dfrac{5}{2}+(-\dfrac{1}{2})(2-1)=2;\\a_3=\dfrac{5}{2}+(-\dfrac{1}{2})(3-1)=\dfrac{3}{2};\\a_4=\dfrac{5}{2}+(-\dfrac{1}{2})(4-1)=1; \\a_5=\dfrac{5}{2}+(-\dfrac{1}{2})(5-1)=\dfrac{1}{2};\\a_6=\dfrac{5}{2}+(-\dfrac{1}{2})(6-1)=0$ Hence, the first six terms are: $\dfrac{5}{2},2,\dfrac{3}{2},1,\dfrac{1}{2},0$
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