Chapter 9.3, Problem 92E

(a)

To determine

To Find: The figure that determines terms of sequence given by $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ and shows terms of the sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ .

Answer to Problem 92E

The figure for $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ is (i) and for $a_n=20{\left(\frac{4}{3}\right)}^{n-1}$ is (ii).

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

  Figure 1

Calculation:

The given sequence $a_n=20{\left(\frac{4}{3}\right)}^n$ shows the increasing term as

  $r>1$ and this sequence represent graph (ii). The sequence $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ shows that the decreeing term the reason for this is $r<1$ and this represented by (ii)

(b)

To determine

To Find: The infinite sequence that has the term that can be summed.

Answer to Problem 92E

The series that can be summed is $a_n=20{\left(\frac{3}{4}\right)}^{n-1}$ as it satisfied the condition $\left|r\right|<1$

Explanation of Solution

Consider that the finite geometric sequence of the condition $\left|r\right|<1$ can be summed.

In the given problem $r=\frac{3}{4}$ , it is less than 1.

Thus, the series that can actually be summed is,

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