Chapter 8, Problem 17CR

To determine

Find the $n\text{th}$ term of the given geometric sequence and graph the first six terms.

Answer to Problem 17CR

The $n\text{th}$ term $a_n=1{\left(\pm 4\right)}^{n-1}$

Explanation of Solution

Given:

The given datais $a_3=16,a_5=256$ .

Calculation:

The general rule for the $n\text{th}$ term is $a_n=a_1{r}^{n-1}$ .

Where $a_1\to$ first term, $r\to$ common ratio, $n\to$ number of terms.

For $3\text{rd}$ term

  $\begin{array}{l}{a}_3=a_1{\left(r\right)}^{3-1}\\ \Rightarrow 16=a_1{r}^2---\left(i\right)\end{array}$

For $5th$ term.

  $\begin{array}{l}{a}_5=a_1{\left(r\right)}^{5-1}\\ \Rightarrow 256=a_1{r}^4---\left(ii\right)\end{array}$

Divide $\left(ii\right)$ by $\left(i\right)$ .

  $\begin{array}{l}\frac{256}{16}=\frac{a_1{r}^4}{a_1{r}^2}\\ \Rightarrow 16=r^2\\ \Rightarrow \pm \sqrt{16}=r\\ \Rightarrow \pm 4=r\end{array}$

Substitute the value of $r=\pm 4$ in equation $\left(i\right)$ .

  $\begin{array}{l}16=a_1{\left(\pm 4\right)}^2\\ \Rightarrow 16=16a_1\\ \Rightarrow 1=a_1\end{array}$

The $n\text{th}$ term is $a_n=1{\left(\pm 4\right)}^{n-1}$

The first $\text{6}$ terms are

For $r=4$ .

  $\begin{array}{l}{a}_1=1\\ a_2=1\cdot 4=4\\ a_3=4\cdot 4=16\\ a_4=16\cdot 4=64\\ a_5=64\cdot 4=256\\ a_6=256\cdot 4=1024\end{array}$

For $r=-4$

  $\begin{array}{l}{a}_1=1\\ a_2=1\cdot \left(-4\right)=-4\\ a_3=-4\cdot \left(-4\right)=16\\ a_4=16\cdot \left(-4\right)=-64\\ a_5=-64\cdot \left(-4\right)=256\\ a_6=256\cdot \left(-4\right)=-1024\end{array}$

Graphs of the first six terms are given below.

  

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