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: 36


$r=\dfrac{t}{2}$ $a_5=\dfrac{t^5}{16}$ The $n^{th}$ term of the geometric sequence is: $a_n=t \cdot \left(\dfrac{t}{2}\right)^{n-1}$.

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: $\require{cancel} r=\dfrac{\frac{t^2}{2}}{t} \\r=\dfrac{t^2}{2} \cdot \dfrac{1}{t} \\r=\dfrac{\cancel{t^2}t}{2} \cdot \dfrac{1}{\cancel{t}} \\r=\dfrac{t}{2}$ The fifth term can be found by multiplying the common ratio to the fourth term. The fourth term is $\dfrac{t^4}{8}$. Thus, the fifth term is: $a_5=\dfrac{t^4}{8} \cdot \dfrac{t}{2}$ Use the rule $a^m \cdot a^n = a^{m+n}$ to obtain: $a_5=\dfrac{t^{4+1}}{16} \\a_5=\dfrac{t^5}{16}$ With a first term of $t$ and a common ratio of $r=\dfrac{t}{2}$, the $n^{th}$ term of the geometric sequence is: $a_n=a \cdot r^{n-1} \\a_n=t \cdot \left(\dfrac{t}{2}\right)^{n-1}$
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