Chapter 11, Problem 5CR

To determine

To calculate:

The $9^{th}$ term of this geometric- sequence $\frac{81}{8},\frac{27}{4},\frac{9}{3},\mathrm{3....}$

Answer to Problem 5CR

The $9^{th}$ term of this geometric- sequence $\frac{81}{8},\frac{27}{4},\frac{9}{3},\mathrm{3....}$ is $t_9=\frac{64}{243}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}{a}_1=\frac{81}{8}\\ a_2=\frac{27}{4}\\ a_3=\frac{9}{2}\\ a_4=3\end{array}$

Formula used:

Common ratio: $r=\frac{a_n}{a_{n-1}}$

Arithmetic progression: $t_n=a{r}^{n-1}$

Calculation:

Consider the following sequence:

  $\begin{array}{l}r=\frac{a_2}{a_1}=\frac{81/8}{27/4}=1.5\\ r=\frac{a_3}{a_2}=\frac{27/4}{9/2}=1.5\\ r=\frac{a_4}{a_3}=\frac{9/2}{3}=1.5\end{array}$

Apply the formula of arithmetic progression:

  $t_n=a{r}^{n-1}$

Put the values in the formula:

  $\begin{array}{l}{t}_9=a{r}^{9-1}\\ t_9=\frac{81}{8}\times {\left(\frac{3}{2}\right)}^8\\ t_9=\frac{81}{8}\times \frac{6561}{256}\\ t_9=\frac{64}{243}\end{array}$

The $9^{th}$ term of this geometric- sequence $\frac{81}{8},\frac{27}{4},\frac{9}{3},\mathrm{3....}$ is $t_9=\frac{64}{243}$ .