Chapter 8.5, Problem 24E

To determine

To write: a recursive rule for the sequence shown in the given graph.

Answer to Problem 24E

  $a_n=\frac{1}{2}{a}_{n-1}$ .

Explanation of Solution

Given:

The given graph is shown below:

  

Calculation:

According to the given graph, the sequence is $8,4,2,1,\mathrm{......}$ .

Therefore, the terms of the given sequence are $a_1=8,a_2=4,a_3=2,\mathrm{.....}\text{ and so on}$ .

Now, from the given terms, it is found that

  $\begin{array}{l}{a}_1=8\{\text{given terms}}\\ a_n=\frac{1}{2}{a}_{n-1}\end{array}$

Now, check this regressive rule for 2^nd and 3^rd term and so on.

  $\begin{array}{l}{a}_n=\frac{1}{2}{a}_{n-1}\{\text{recursive rule}}\\ a_2=\frac{1}{2}{a}_{2-1}\{\text{put }n=2\text{}}\\ a_2=\frac{1}{2}{a}_1\\ a_2=\frac{1}{2}\left(8\right)\{\text{given }{a}_1\text{=8}}\\ a_2=4\{\text{true}}\end{array}$

  $\begin{array}{l}{a}_3=\frac{1}{2}{a}_{3-1}\{\text{put }n=3\text{}}\\ a_3=\frac{1}{2}{a}_1\\ a_3=\frac{1}{2}\left(8\right)\{\text{given }{a}_2=4\}\\ a_3=4\{\text{true}}\end{array}$

Hence, the recursive rule for the given sequence is $a_n=\frac{1}{2}{a}_{n-1}$ .