College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.3 - Geometric Sequences - 8.3 Exercises: 35

Answer

$r=3^{2/3}$ $a_5=3^{11/3}$ The $n^{th}$ term of the geometric sequence is: $a_n=3 \cdot \left(3^{2/3}\right)^{n-1}$

Work Step by Step

RECALL: (1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms: $r =\dfrac{a_n}{a_{n-1}}$ (2) The $n^{th}$ term of a geometric sequence is given by the formula: $a_n = a\cdot r^{n-1}$ where $a$ = first term $r$ = common ratio The sequence is said to be geometric. Thus, we can proceed to solving for the common ratio: $r=\dfrac{3^{5/3}}{3}$ Use the rule $\dfrac{a^m}{a^n} = a^{m-n}$ to obtain: $r = 3^{5/3-1} \\r=3^{5/3 - 3/3} \\r=3^{2/3}$ The fifth term can be found by multiplying the common ratio to the fourth term. The fourth term is $27$. Thus, the fifth term is: $a_5=27 \cdot \left(3^{2/3}\right) \\a_5=3^3 \cdot 3^{2/3}$ Use the rule $a^m\cdot a^n = a^{m+n}$ to obtain: $a_5=3^{3+2/3} \\a_5=3^{9/3 + 2/3} \\a_5=3^{11/3}$ With a first term of $3$ and a common ratio of $r=3^{2/3}$, the $n^{th}$ term of the geometric sequence is: $a_n=a \cdot r^{n-1} \\a_n=3 \cdot \left(3^{2/3}\right)^{n-1}$
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