Chapter 8.3, Problem 122E

(a)

To determine

To find: which figure shows the terms of the sequence $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ , and which shows the terms of $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ .

Answer to Problem 122E

The terms of the sequence $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ is (ii)

The terms of the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ is (i).

Explanation of Solution

Given information:

Given figure

  

Calculation:

In the sequence $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ , the value of the expression $\frac{4}{3}>1$ . Thus, the value of ${\left(\frac{4}{3}\right)}^{n-1}$ as well as $20{\left(\frac{4}{3}\right)}^{n-1}$ increases as $n$ increases. So, the figure that shows the terms of the sequence $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ is (ii).

In sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ , the value of the expression $\frac{3}{4}<1$ . Thus, the value of ${\left(\frac{3}{4}\right)}^{n-1}$ as well as $20{\left(\frac{3}{4}\right)}^{n-1}$ decreases as $n$ increases. So, the figure that shows the terms of the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ is (i).

(b)

To determine

To find: the sequence has terms can be summed.

Answer to Problem 122E

The terms can be summed for the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ .

Explanation of Solution

Given information:

Given figure

  

Calculation:

To find the sum of a geometric series, modulus of $r$ should be less than 1. That is,

  $\left|r\right|<1$

For the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ ,

  $\begin{array}{c}\left|r\right|=\frac{3}{4}\\ <1\end{array}$

Thus, the terms can be summed for the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ .