Chapter 8.1, Problem 15E

(a)

To determine

To find the common ratio.

Answer to Problem 15E

The common ratio is $\frac{1}{2}$

Explanation of Solution

Given:

  $8,4,2,1,\dots$

Concept Used:

A sequence $\left\{a_n\right\}$ is a geometric sequence if it can be written in the form $\left\{a,a\cdot r,a\cdot r^2,\dots ,a\cdot r^{n-1},\dots \right\}$ for some constant $r$ ,the number $r$is the common ratio.

Calculation:

By comparing the geometric sequence:

  $\begin{array}{c}a\cdot r=4,\\ a=8\end{array}$

The common ratio is:

  $\begin{array}{c}r=\frac{a\cdot r}{a}\\ =\frac{4}{8}\\ =\frac{1}{2}\end{array}$

Conclusion:

The common ratio is $\frac{1}{2}$

(b)

To determine

To find: The ninth term.

Answer to Problem 15E

The ninth term is $0.03125$

Explanation of Solution

Given:

Sequence; $8,4,2,1,\dots$

Concept Used:

• The $n^{th}$ term is:

$a_n=a\cdot r^{n-1}\text{for}\text{all n}\ge \text{2}$

Where, $a$ =First term, $r$ =common ratio and $n$ = $n^{th}$ term.

Calculation:

Here,

  $a=8$ ,

  $r=\frac{1}{2}$

Ninth term is:

  $\begin{array}{c}{a}_9=8\cdot {\left(\frac{1}{2}\right)}^8\\ =8\left(\frac{1}{256}\right)\\ =0.03125\end{array}$

Conclusion:

The ninth term is $0.03125$ .

(c).

To determine

To find a recursive rule for $n^{th}$ term.

Answer to Problem 15E

A recursive rule for $n^{th}$ term.is $a_n=\left(\frac{1}{2}\right)\cdot a_{n-1}$ .

Explanation of Solution

Given:

Sequence; $8,4,2,1,\dots$

Concept Used:

Each term is a geometric sequence can be obtained recursively from its preceding term by multiplying $r$ :

  $a_n=a_{n-1}\cdot r\text{for}\text{all}n\ge 2$

Calculation:

Here,

  $a=8$ ,

  $r=\frac{1}{2}$

By using recursive rule, $n^{th}$ is:

  $\begin{array}{c}{a}_n=r\cdot a_{n-1}\\ =\left(\frac{1}{2}\right)\cdot a_{n-1}\end{array}$

Conclusion:

A recursive rule for $n^{th}$ term.is $a_n=\left(\frac{1}{2}\right)\cdot a_{n-1}$ .

(d).

To determine

To find an explicit rule for $n^{th}$ term.

Answer to Problem 15E

An explicit rule for $n^{th}$ term is $a_n=8\cdot {\left(\frac{1}{2}\right)}^{n-1}$ .

Explanation of Solution

Given:

  $8,4,2,1,\dots$

Concept Used:

The rule for explicit $n^{th}$ is:

  $a_n=a\cdot {\left(r\right)}^{n-1}$

Where, $a$ =First term, $r$ =common ratio and $n$ = $n^{th}$ term.

Calculation:

Here,

  $a=8$ ,

  $r=\frac{1}{2}$

The explicit rule is:

  $\begin{array}{c}{a}_n=a\cdot r^{n-1}\\ =8\cdot {\left(\frac{1}{2}\right)}^{n-1}\end{array}$

Conclusion:

An explicit rule for $n^{th}$ term is $a_n=8\cdot {\left(\frac{1}{2}\right)}^{n-1}$ .