## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 12 - Section 12.3 - Geometric Sequences - 12.3 Exercises - Page 865: 38

#### Answer

$r=5^{c}$ $a_{5}=5^{4c+1}$ $a_{n}=5^{(n-1)c+1}$

#### Work Step by Step

A geometric sequence is a sequence whose terms are obtained by multiplying each term by the same fixed constant $r$ to get the next term. A geometric sequence has the form $a, ar, ar^{2}, ar^{3}, \ldots$ The number $a$ is the first term of the sequence, and the number $r$is the common ratio. The nth term of the sequence is $\quad a_{n}=ar^{n-1}$ -------------- The common ratio is the ratio between any two neighboring terms: $r=\displaystyle \frac{5^{c+1}}{5^{1}}=5^{c+1-1}=5^{c}$ The first term $a=5.$ The fifth term (n=5) $a_{5}=5(5^{c})^{5-1}=5(5^{4c})=5^{4c+1}$ The general, nth term: $a_{n}=5(5^{c})^{n-1}=5(5^{nc-c})=5^{nc-c+1}$

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