Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.3 - Geoetric Sequences and Series - Exercise Set - Page 1076: 105


False. The sequence is not geometric.

Work Step by Step

A series is said to be in geometric progression if the common ratio is the same, as shown below: $\begin{align} & \frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{a}_{3}}}{{{a}_{2}}} \\ & =\frac{{{a}_{4}}}{{{a}_{3}}} \end{align}$ Here ${{a}_{1}}=2,{{a}_{2}}=6,{{a}_{3}}=24,{{a}_{4}}=120$. So; $\begin{align} & \frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{a}_{3}}}{{{a}_{2}}} \\ & =\frac{{{a}_{4}}}{{{a}_{3}}} \\ & \frac{6}{2}=\frac{24}{6} \\ & =\frac{120}{24} \end{align}$ It can be further solved as below: $\begin{align} & 3\ne 4 \\ & \ne 5 \end{align}$ Hence, the given statement is false. For the statement to be true, $\begin{align} & \frac{{{a}_{2}}}{{{a}_{1}}}=\frac{{{a}_{3}}}{{{a}_{2}}} \\ & =\frac{{{a}_{4}}}{{{a}_{3}}} \end{align}$
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