College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 17

Answer

The sequence is geometric with a common ratio of $\dfrac{3}{2}$. The first four terms are: $t_1=\dfrac{1}{2}$ $t_2=\dfrac{3}{4}$ $t_3=\dfrac{9}{8}$ $t_4=\dfrac{27}{16}$

Work Step by Step

Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula: $t_1=\dfrac{3^{1-1}}{2^1}=\dfrac{3^0}{2}=\dfrac{1}{2}$ $t_2=\dfrac{3^{2-1}}{2^2}=\dfrac{3^1}{4}=\dfrac{3}{4}$ $t_3=\dfrac{3^{3-1}}{2^3}=\dfrac{3^2}{8}=\dfrac{9}{8}$ $t_4=\dfrac{3^{4-1}}{2^4}=\dfrac{3^3}{16}=\dfrac{27}{16}$ To check if the sequence is geometric, find the ratio of each pair of successive terms: $\dfrac{a_2}{a_1}=\dfrac{\frac{3}{4}}{\frac{1}{2}}=\dfrac{3}{4} \cdot \dfrac{2}{1}=\dfrac{3}{2}$ $\dfrac{a_3}{a_2}=\dfrac{\frac{9}{8}}{\frac{3}{4}}=\dfrac{9}{8} \cdot \dfrac{4}{3}=\dfrac{3}{2}$ $\dfrac{a_4}{a_3}=\dfrac{\frac{27}{16}}{\frac{9}{8}}=\dfrac{27}{16} \cdot \dfrac{8}{9}=\dfrac{3}{2}$ The common ratios are the same, so the sequence is geometric with a common ratio of $\frac{3}{2}$.
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