Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.2 Arithmetic Sequences and Series - 14.2 Exercise Set - Page 902: 30

Answer

$d=0.5$ $a_1=\dfrac{1}{3}$ $a_2=\dfrac{5}{6}$ $a_3=\dfrac{4}{3}$ $a_4=\dfrac{11}{6}$ $a_5=\dfrac{7}{3}$

Work Step by Step

We have an arithmetic sequence in which $a_{17}=\dfrac{25}{3}$ and $a_{32}=\dfrac{95}{6}.$ We have to find first five terms and $a_1,$ $d.$ In order to find $a_{1}$ and $d$ we will use the fact that any term of arithmetic sequence is expressed through the formula $a_{n}=a_1+d(n-1)$ Hence, $a_{17}=a_1+d\cdot16$ $a_{32}=a_1+d\cdot31$ $a_{17}-a_{32}=a_1+d\cdot16-a_1-d\cdot31=d(16-31)=-15d$ $\dfrac{25}{3}-\dfrac{95}{6}=-15d$ $-\dfrac{45}{6}=-15d$ And $d=0.5$ Now we can use any equation to find $a_1.$ $\dfrac{25}{3}=a_1+0.5\cdot16$ $\dfrac{25}{3}-8=a_1$ $a_1=\dfrac{1}{3}$ The first five terms will be $a_1=\dfrac{1}{3}$ $a_2=a_1+d=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}$ $a_3=a_1+2d=\dfrac{1}{3}+1=\dfrac{4}{3}$ $a_4=a_1+3d=\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{11}{6}$ $a_5=a_1+4d=\dfrac{1}{3}+2=\dfrac{7}{3}$
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