Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.2 Arithmetic Sequences - 11.2 Assess Your Understanding - Page 835: 51



Work Step by Step

We can see that there are $100$ terms and these terms are part of an arithmetic sequence, so we have: $a_1=(6)-(1/2)(1)=\dfrac{11}{2} \\ a_{100}=6-(\dfrac{1}{2})(100)=-44$ There is a constant difference between the terms of: $d=a_{n+1}-a_n=6-\dfrac{1}{2}(n+1) -(6-\dfrac{n}{2})=\dfrac{-1}{2}$ The sum of the first $n$ terms of an arithmetic sequence is given by: $S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right) ..(1)$ Now, we plug in the above data into Equation-1 to obtain: $S_{100}= \dfrac{100}{2}[\dfrac{11}{2}+(-44)] \\=(-25)(77) \\= -1925$ Therefore, the sum of the arithmetic sequence is: $-1925$.
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