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 $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ without calculation.

Answer to Problem 122E

Graph (i) corresponds to $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ because the common ration is $\frac{3}{4}<1$ and the graph is decreasing.

Graph (ii) corresponds to $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ because the common ration is $\frac{4}{3}>1$ and the graph is increasing.

Explanation of Solution

Given information: Given figures are given below.

  

Calculation:

Graph (i) corresponds to $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ because the common ration is $\frac{3}{4}<1$ and the graph is decreasing.

Graph (ii) corresponds to $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ because the common ration is $\frac{4}{3}>1$ and the graph is increasing.

b.

To determine

To find: which sequence has terms that can be summed, explain.

Answer to Problem 122E

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

Because common ration is $0<\left|\frac{3}{4}\right|<1$ .

Explanation of Solution

Calculation:

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

Because common ration is $0<\left|\frac{3}{4}\right|<1$ .