Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 50



Work Step by Step

Here, we have $a_n= a_1+d(n-1)$ for the Arithmetic series. The common difference between the successive terms is $d=6.5$ and $a_5=92$ Here, we have $a_5= a_1-(5-1)d$ or, $a_5=a_1+4d \implies a_1=92-4(6.5)=66$ Now, $a_n=66+(6.5) \times (n-1)=6.5 n+59.5$ Hence, $a_n=6.5n+59.5$
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