Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises - Page 737: 15

Answer

$a_n=625\left(\frac{1}{5}\right)^{n-1}$

Work Step by Step

$625, 125, 25, 5, 1, 0.2, . . .$ $\frac{a_2}{a_1}=\frac{125}{625}=\frac{1}{5}=r$ $ra_2=\frac{1}{5}\times125=25=a_3$ $ra_3=\frac{1}{5}\times25=5=a_4$ and so on. The sequence is geometric with ratio $r=\frac{1}{5}$. The $n^{th}$ term is given by $a_n=625\left(\frac{1}{5}\right)^{n-1}$

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