Algebra 2 (1st Edition)

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

Chapter 12 Sequences and Series - Standardized Test Practice - Extended Response - Page 847: 20a

Answer

See below

Work Step by Step

At the end of the first month, the equation will be $16000(1+0.75\%)$ The loan at the beginning of the second month is $(1+0.75\%)a_1-300=1.0075a_1-300$ The loan at the beginning of the third month is $(1+0.75\%)a_2-300=1.0075a_2-300$ The loan at the beginning of the nth month is $a_n=1.0075a_{n-1}-300$ with $n \geq 2$
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