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 834: 48

Answer

$4901$.

Work Step by Step

The nth term of the arithmetic sequence is given by: $a_n=a_1+(n-1) d \\ 50=8+\dfrac{1}{4}(n-1) \\168 =(n-1) \\ n= 169$ We see that there is a constant difference between the terms of $d=\dfrac{1}{4}$ and the terms are part of an arithmetic sequence. The terms of the sum are the first $193$ terms of an arithmetic sequence, starting with $a_{1}=8$ and with a difference of $d=\dfrac{1}{4}$. The sum of the first $n$ terms of an arithmetic sequence is given by: $S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right)$ Now, $S_{169}= \dfrac{169}{2}[8+50] \\=(169)(29) \\= 4901$ Therefore, the sum of the arithmetic sequence is: $4901$.
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