Chapter 9.3, Problem 20E

To determine

To write: the first five terms of the geometric sequence.

Answer to Problem 20E

The first five terms of the geometric sequence are $4,-2\sqrt{2},2,-\sqrt{2},1$ .

Explanation of Solution

Given information:

The given sequence,

  $a_1=4,r=-1/\sqrt{2}$

Concept used:

The $n$ th term of geometric sequence has the form $a_1=a_1{r}^{n-1}$ where $r$ is the common ratio of the consecutive terms of the sequence.

Every geometric sequence can be written in the following form.

  $a_1=a_1,a_2=a_{1}r,a_3=a_1{r}^2,a_4=a_1{r}^3,\mathrm{...},a_n=a_1{r}^{n-1},\mathrm{..}$

Calculation:

The common ratio and the first term of the sequence is given

  $\begin{array}{l}{a}_1=4\\ r=-\frac{1}{\sqrt{2}}\end{array}$

To find the second term, multiply first term $\pi$ as shown below.

  $\begin{array}{c}{a}_2=a_1{r}^{2-1}\\ =4\left(-\frac{1}{\sqrt{2}}\right)\\ =-2\sqrt{2}\end{array}$

The 3^rd term can be calculated as

  $\begin{array}{c}{a}_3=a_1{r}^{2-1}\\ =4{\left(-\frac{1}{\sqrt{2}}\right)}^2\\ =2\end{array}$

The 4^th term can be calculated as

  $\begin{array}{c}{a}_4=a_1{r}^{2-1}\\ =4{\left(-\frac{1}{\sqrt{2}}\right)}^3\\ =-\sqrt{2}\end{array}$

The 5^th term can be calculated as

  $\begin{array}{c}{a}_5=a_1{r}^{2-1}\\ =4{\left(-\frac{1}{\sqrt{2}}\right)}^4\\ =1\end{array}$

Hence, the first five terms of the geometric sequence are

  $4,-2\sqrt{2},2,-\sqrt{2},1$ .