Chapter 9.3, Problem 52E

To determine

ToGraph:The first $10$ terms of the sequence $a_n=18{\left(0.7\right)}^{n-1}$ , using a graphing utility.

Answer to Problem 52E

The graph of the first $10$ terms of the given sequence is

  

Explanation of Solution

Given:

The sequence $a_n=18{\left(0.7\right)}^{n-1}$

Concept Used:

To find the first $10$ terms of a sequence. Substitute the integral values of $n$ from $n=1$ to $n=10$ into the sequence and calculate the value of the sequence at each value of $n$

Calculation:

For the givensequence $a_n=18{\left(0.7\right)}^{n-1}$

For $n=1$

  $\begin{array}{c}{a}_1=18{\left(0.7\right)}^{1-1}\\ =18{\left(0.7\right)}^0\\ =18\end{array}$

For $n=2$

  $\begin{array}{c}{a}_2=18{\left(0.7\right)}^{2-1}\\ =18{\left(0.7\right)}^1\\ =12.6\end{array}$

For $n=3$

  $\begin{array}{c}{a}_3=18{\left(0.7\right)}^{3-1}\\ =18{\left(0.7\right)}^2\\ =8.82\end{array}$

For $n=4$

  $\begin{array}{c}{a}_4=18{\left(0.7\right)}^{4-1}\\ =18{\left(0.7\right)}^3\\ =6.174\end{array}$

For $n=5$

  $\begin{array}{c}{a}_5=18{\left(0.7\right)}^{5-1}\\ =18{\left(0.7\right)}^4\\ =4.3218\end{array}$

For $n=6$

  $\begin{array}{c}{a}_6=18{\left(0.7\right)}^{6-1}\\ =18{\left(0.7\right)}^5\\ =3.02526\end{array}$

For $n=7$

  $\begin{array}{c}{a}_7=18{\left(0.7\right)}^{7-1}\\ =18{\left(0.7\right)}^6\\ =\text{2}\text{.117682}\end{array}$

For $n=8$

  $\begin{array}{c}{a}_8=18{\left(0.7\right)}^{8-1}\\ =18{\left(0.7\right)}^7\\ =1.4823774\end{array}$

For $n=9$

  $\begin{array}{c}{a}_9=18{\left(0.7\right)}^{9-1}\\ =18{\left(0.7\right)}^8\\ =1.03766418\end{array}$

For $n=10$

  $\begin{array}{c}{a}_{10}=18{\left(0.7\right)}^{10-1}\\ =18{\left(0.7\right)}^9\\ =0.726364926\end{array}$

Plotting the first $10$ terms of the given sequence, we have

  

Conclusion:

The graph of the first $10$ terms of the given sequence is