Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 12 Sequences and Series - 12.2 Analyze Arithmetic Sequences and Series - 12.2 Exercises - Skill Practice - Page 806: 18


$a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$ ; $a_{20}=\dfrac{38}{3}$

Work Step by Step

We know that $a_n= a_1+(n-1) d$ Here, we have $d=\dfrac{2}{3}-0=\dfrac{2}{3}$ $a_n= 0+(n-1) \times (\dfrac{2}{3})$ This gives: $a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$ $a_{20}=-\dfrac{2}{3}+\dfrac{2}{3} \times 20=\dfrac{38}{3}$ Hence, $a_n=-\dfrac{2}{3}+\dfrac{2}{3}n$ ; $a_{20}=\dfrac{38}{3}$
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