Chapter 9, Problem 45RE

To determine

To find :The first five terms of the geometric sequence.

Answer to Problem 45RE

The first five terms of the geometric sequence are

  $a_1=9,a_2=6,a_3=4,a_4=\frac{8}{3},a_5=\frac{16}{9}$ or $a_1=9,a_2=-6,a_3=4,a_4=-\frac{8}{3},a_5=\frac{16}{9}$

Explanation of Solution

Given information: $a_1=9,a_3=4$

Calculation:

The first term and geometric sequence is $a_1=9,a_3=4$ respectively.

The $n$ term of geometric sequence is

  $a_n=a_1{r}^{n-1}$

Therefore,

  $\begin{array}{l}{a}_3=a_1{r}^{3-1}\\ 4=9r^2\\ r^2=\frac{4}{9}\\ r=\pm \frac{2}{3}\end{array}$

Now, if $r=\frac{2}{3}$

By substituting $n=2,3,4,5$ in the above equation, we get the second, third ,fourth and fifth term of the sequence.

  $\begin{array}{l}{a}_2=a_1{r}^{2-1}=a_{1}r=9\times \left(\frac{2}{3}\right)=6\\ a_3=a_1{r}^{3-1}=a_1{r}^2=9\times {\left(\frac{2}{3}\right)}^2=4\\ a_4=a_1{r}^{4-1}=a_1{r}^3=9\times {\left(\frac{2}{3}\right)}^3=\frac{8}{3}\\ a_5=a_1{r}^{5-1}=a_1{r}^4=9\times {\left(\frac{2}{3}\right)}^4=\frac{16}{9}\end{array}$

If $r=-\frac{2}{3}$ , then the second, third ,fourth and fifth term of the sequence are

  $\begin{array}{l}{a}_2=a_1{r}^{2-1}=a_{1}r=9\times \left(-\frac{2}{3}\right)=-6\\ a_3=a_1{r}^{3-1}=a_1{r}^2=9\times {\left(-\frac{2}{3}\right)}^2=4\\ a_4=a_1{r}^{4-1}=a_1{r}^3=9\times {\left(-\frac{2}{3}\right)}^3=-\frac{8}{3}\\ a_5=a_1{r}^{5-1}=a_1{r}^4=9\times {\left(-\frac{2}{3}\right)}^4=\frac{16}{9}\end{array}$