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: 19


$a_n=\dfrac{7}{3}-\dfrac{1}{3}n$ ; $a_{20}=\dfrac{-13}{3}$

Work Step by Step

We know that $a_n= a_1+(n-1) d$ Here, we have $d=\dfrac{5}{3}-2=-\dfrac{1}{3}$ $a_n= 2+(n-1) \times (\dfrac{-1}{3})$ This gives: $a_n=\dfrac{7}{3}-\dfrac{1}{3}n$ $a_{20}=\dfrac{7}{3}-\dfrac{1}{3} \times 20=\dfrac{-13}{3}$ Hence, $a_n=\dfrac{7}{3}-\dfrac{1}{3}n$ ; $a_{20}=\dfrac{-13}{3}$
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