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: 30

Answer

$r=\dfrac{2}{3}$ $a_5=\dfrac{112}{81}$ The $n^{th}$ term of the geometric sequence is: $a_n=7 \cdot \left(\dfrac{2}{3}\right)^{n-1}$

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: $r=\dfrac{\frac{14}{3}}{7} \\r = \dfrac{14}{3} \cdot \dfrac{1}{7} \\r=\dfrac{2}{3}$ The fifth terms can be found by multiplying the common ratio to the fourth term. The fourth term is $\dfrac{56}{27}$. Thus, the fifth term is: $a_5=\dfrac{56}{27} \cdot \dfrac{2}{3} \\a_5=\dfrac{112}{81}$ With a first term of $7$ and a common ratio of $r=\dfrac{2}{3}$, the $n^{th}$ term of the geometric sequence is: $a_n=a \cdot r^{n-1} \\a_n=7 \cdot \left(\dfrac{2}{3}\right)^{n-1}$
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