 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q52.

Expert-verified Found in: Page 596 ### Algebra 1

Book edition Student Edition
Author(s) Carter, Cuevas, Day, Holiday, Luchin
Pages 801 pages
ISBN 9780078884801 # Find the next three terms in each geometric sequence. $\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{.}\mathbf{....}$

The next three terms in the geometric sequence are $\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{1}\mathbf{,}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\mathbf{1}$.

See the step by step solution

## Step 1. State the concept used.

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3.

The ration of ${2}^{nd}$ term and first term is called the common ratio.

## Step 2. Find the common ratio.

The sequence: $-1,1,-1,1,.....$

Each term in a geometric sequence can be expressed in terms of the first term ${a}_{1}$ and the common ratio $r$. Since each succeeding term is formulated from one or more previous terms, this is a recursive formula.

First term ${a}_{1}=-1$ and the common ratio is:

$\begin{array}{c}r=\text{\hspace{0.17em}\hspace{0.17em}}\frac{{a}_{2}}{{a}_{1}}\\ =\frac{1}{\left(-1\right)}\\ =-1\end{array}$

## Step 3. Find the next three terms.

In order to calculate the next three terms of the geometric sequence substitute $n=4,5,6$ and $-1$ for ${a}_{1}$ into the formula ${a}_{n}={a}_{1}{r}^{n-1}$.

 Terms Symbol In terms of ${a}_{1}$ and $r$ Numbers Fourth term ${a}_{4}$ ${a}_{1}\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}{r}^{3}$ $\left(-1\right){\left(-1\right)}^{3}=\left(-1\right)×\left(-1\right)=1$ Fifth term ${a}_{5}$ ${a}_{1}\cdot {r}^{4}$ $\left(-1\right){\left(-1\right)}^{4}=\left(-1\right)×\left(1\right)=-1$ Sixth term ${a}_{6}$ ${a}_{1}\cdot {r}^{5}$ $\left(-1\right){\left(-1\right)}^{5}=\left(-1\right)×\left(-1\right)=1$

Thus, next three terms of the sequence $-1,1,-1,1,.....$ are $-1,\text{\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}\hspace{0.17em}}-1$. ### Want to see more solutions like these? 