College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.2 - Page 727: 86

Answer

The given $n^{th}$ term ( $a_{n}$) = $3.5^{n}$ $\frac{a_{2}}{a_{1}}$ = $\frac{12.25}{3.5}$ = 3.5 $\frac{a_{3}}{a_{2}}$ = $\frac{42.875}{12.25}$ = 3.5 $\frac{a_{4}}{a_{3}}$ = $\frac{150.0625}{42.875}$ = 3.5 $\frac{a_{5}}{a_{4}}$ = $\frac{525.21875}{150.0625}$ = 3.5 From this we observe that $\frac{a_{2}}{a_{1}}$ = $\frac{a_{3}}{a_{2}}$ = $\frac{a_{4}}{a_{3}}$ = $\frac{a_{5}}{a_{4}}$ = 3.5 Here each term after the first term is found by multiplying the previous one by a fixed, non-zero number called the common ratio = 3.5.

Work Step by Step

The given $n^{th}$ term ( $a_{n}$) = $3.5^{n}$ $a_{1}$ = $3.5^{1}$ = 3.5 $a_{2}$ = $3.5^{2}$ = 12.25 $a_{3}$ = $3.5^{3}$ = 42.875 $a_{4}$ = $3.5^{4}$ = 150.0625 $a_{5}$ = $3.5^{5}$ = 525. 21875 $\frac{a_{2}}{a_{1}}$ = $\frac{12.25}{3.5}$ = 3.5 $\frac{a_{3}}{a_{2}}$ = $\frac{42.875}{12.25}$ = 3.5 $\frac{a_{4}}{a_{3}}$ = $\frac{150.0625}{42.875}$ = 3.5 $\frac{a_{5}}{a_{4}}$ = $\frac{525.21875}{150.0625}$ = 3.5 From this we observe that $\frac{a_{2}}{a_{1}}$ = $\frac{a_{3}}{a_{2}}$ = $\frac{a_{4}}{a_{3}}$ = $\frac{a_{5}}{a_{4}}$ = 3.5 Here each term after the first term is found by multiplying the previous one by a fixed, non-zero number called the common ratio = 3.5.
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