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: 52

Answer

$1120$.

Work Step by Step

We can see that there are $80$ terms and these terms are part of an arithmetic sequence, so we have: $a_1=(\dfrac{1}{3})(1)+\dfrac{1}{2}=\dfrac{5}{6} \\ a_{80}=(\dfrac{1}{3})(80)+\dfrac{1}{2}=\dfrac{163}{6}$ There is a constant difference between the terms of: $d=a_{n+1}-a_n=(\dfrac{1}{3})(n+1)+\dfrac{1}{2}-(\dfrac{1}{3}n+\dfrac{1}{2})=\dfrac{1}{3}$ 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: $S_{80}= \dfrac{80}{2}[\dfrac{5}{6}+\dfrac{163}{6}] \\=(40)(28) \\= 1120$ Therefore, the sum of the arithmetic sequence is: $1120$.
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