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 - Page 615: 19


The ratios are difference so the sequence is not geometric.

Work Step by Step

A geometric sequence has a common ratio $r$. The common ratio is multiplied to the current term to get the next term of the sequence. The common ratio is equal to the the quotient of a term and the term before it. Solve for the ratio of each pair of consecutive terms to obtain: $\dfrac{\frac{1}{3}}{\frac{1}{2}} = \dfrac{1}{3} \cdot \dfrac{2}{1}=\dfrac{2}{3} \\\dfrac{\frac{1}{4}}{\frac{1}{3}}=\dfrac{1}{4} \cdot \dfrac{3}{1}= \dfrac{3}{4} \\\dfrac{-1}{3} = -\dfrac{1}{3} $ Since the ratios are different, there is no need to find the ratio of the last pair. The sequence is not geometric.
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