Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.3 Geometric Sequences; Geometric Series - 11.3 Assess Your Understanding - Page 845: 83

Answer

$x=-4$

Work Step by Step

A sequence is geometric if there exists a common ratio $r$ among consecutive terms such as: $r=\dfrac{a_n}{a_{nāˆ’1}}$. In order for a sequence to be geometric, the quotient of all consecutive terms must remain constant. So, we have: $\dfrac{a_3}{a_2}=\dfrac{a_2}{a_1}$. Substituting the given values of the first three terms, we obtain: $$\dfrac{x+3}{x+2}=\dfrac{x+2}{x}$$ Next, we will do cross-multiplication to obtain: \begin{align*}\dfrac{x+3}{x+2}&=\dfrac{x+2}{x}\\x(x+3)&=(x+2)(x+2)\\ x^2+3x&=x^2+4x+4\\4x+4&=3x\\ 4x+4-3x&=3x-3x\\ x+4&=0\\ x&=-4\end{align*}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.