Chapter 9.3, Problem 45E

To determine

To find: the 6th term of the geometric sequence.

Answer to Problem 45E

The 6th term of the geometric series will be $a_6=-2$ .

Explanation of Solution

Given information:

Two terms of the sequence,

  $a_4=-18,a_7=\frac{2}{3}$

Calculation:

The general term of geometric sequence is given by formula

  $a_n=a_1{r}^{n-1}$

Here, $a_1$ is the first term and $r$ is common ratio.

Now $a_4=a_1{r}^{4-1}$

  $\begin{array}{l}{a}_4=a_1{r}^3\\ -18=a_1{r}^3\mathrm{...............}(1)\end{array}$

And

  $\begin{array}{l}{a}_7=a_1{r}^{7-1}\\ \frac{2}{3}=a_1{r}^6\mathrm{............}(2)\end{array}$

Now, substitute the value of $r^3$ from equation (1) to equation (2) to obtain the value of $a_1$ .

  $\begin{array}{l}\frac{2}{3}=a_1{\left(r^3\right)}^2\\ \frac{2}{3}=a_1{\left(-\frac{18}{a_1}\right)}^2\\ \frac{2}{3}=\frac{324}{a_1}\\ a_1=486\end{array}$

Now, substitute the value $a_1$ in equation (1), to obtain the value of $r$ .

  $\begin{array}{c}-18=a_1{r}^3\\ -18=486\left(r^3\right)\\ r^3=-\frac{18}{486}\end{array}$

Now, calculate the value of $a_6$ by using the formula as shown:

  $\begin{array}{c}{a}_6=a_1{r}^5\\ a_6=486{\left(-\frac{1}{27}\right)}^{\frac{5}{3}}\\ a_6=486{\left(-\frac{1}{3^3}\right)}^{\frac{5}{3}}\\ a_6=-\frac{486}{243}\\ a_6=-2\end{array}$

Hence, 6^thterm will be $a_6=-2$ .