Chapter 9.3, Problem 27E

Solution

a.

To calculate: The common ratio of the sequence $1,-2,4,-8,16,\mathrm{...}$

The common ratio of the sequence is $-2$ .

Given information:

The sequence is $1,1,-2,4,-8,16,\mathrm{...}$

Formula used:

The common ratio ( r ) is $r=\frac{a_2}{a_1}$ where $a_2$ is the second term of the sequence and $a_1$ is the first term of the sequence.

Calculation:

Consider the given sequence.

  $1,-2,4,-8,16,\mathrm{...}$

Use the common ratio formula.

  $\begin{array}{c}r=\frac{-2}{1}\\ =-2\end{array}$

The common ratio is $-2$ .

b.

To calculate: The eighth term of the sequence $1,-2,4,-8,16,\mathrm{...}$

The eighth term of the given sequence is $-128$ .

Given information:

The given sequence is $1,-2,4,-8,16,\mathrm{...}$

Formula used:

The formula for the nth term of a geometric sequence is $a_n=a_1{r}^{n-1}$ where $a_1$ is the first term n is the nth term and r is the common ratio.

Calculation:

Consider the given sequence.

The common ratio is $-2$ , the first term is 1 and the value of n is 8.

Substitute the respective values into the formula $a_n=a_1{r}^{n-1}$ .

  $\begin{array}{c}{a}_n=1{\left(-2\right)}^{8-1}\\ =1{\left(-2\right)}^7\\ =-128\end{array}$

The eighth term of the given sequence is $-128$ .

c.

To identify: The recursive rule for the nth term of the sequence $1,-2,4,-8,16,\mathrm{...}$

The recursive rule is $a_1=1,a_n=-2a_{n-1}\text{ for }n\ge 2$ .

Given information:

The given sequence is $1,-2,4,-8,16,\mathrm{...}$

Explanation:

Consider the given sequence.

Let the first term of the sequence is $a_1$ thus the formula for the nth term will be $a_n=a_{n-1}\left(r\right)\text{ for }n\ge 2$ where r is the common ratio.

Since the first term is 1 and the common ratio is $-2$ thus,

  $a_1=1,a_n=-2a_{n-1}\text{ for }n\ge 2$

d.

To identify: An explicit rule for the nth term of the sequence $1,-2,4,-8,16,\mathrm{...}$

The explicit rule is $a_n={\left(2\right)}^{n-1}$ .

Given information:

The given sequence is $1,-2,4,-8,16,\mathrm{...}$

Explanation:

Consider the given sequence.

It is known that the common ratio is $-2$ , and the first term is 1.

Substitute the respective values into the formula $a_n=a_1{r}^{n-1}$ .

  $a_n=1{\left(-2\right)}^{n-1}$

The explicit rule is $a_n={\left(2\right)}^{n-1}$ .