Chapter 8, Problem 4RE

(a)

To determine

The common ratio for the given sequence.

Answer to Problem 4RE

The common difference of $\frac{1}{2},-2,8,-32,\cdots$ is $r=-4$ .

Explanation of Solution

Given: The sequence is $\frac{1}{2},-2,8,-32,\cdots$ .

Definition used:( Geometric Sequence).

A sequence $\left\{a_n\right\}$ is a geometric sequence and it can be written as $\left\{a,a\cdot r,a\cdot r^2,\cdots ,a\cdot r^{n-1},\cdots \right\}$ then the common ratio is r and the nth term is $a_n=a_{n-1}\cdot r\text{ for all }n\ge 2$ .

Calculation:

The terms are $t_1=\frac{1}{2},t_2=-2,t_3=8,t_4=-32,\cdots$

The common ratio between the first two terms is,

  $\begin{array}{c}\frac{t_2}{t_1}=\frac{-2}{\frac{1}{2}}\\ =-4\end{array}$

The common difference between the second term and third term is,

  $\begin{array}{c}\frac{t_3}{t_2}=\frac{8}{-2}\\ =-4\end{array}$ .

Similarly, the common ratio for the consecutive terms of $\frac{1}{2},-2,8,-32,\cdots$ is $r=-4$ .

Therefore, the common difference of $\frac{1}{2},-2,8,-32,\cdots$ is $r=-4$ .

(b)

To determine

The seventh term for the given sequence.

Answer to Problem 4RE

The seventh term of the sequence $\frac{1}{2},-2,8,-32,\cdots$ is $a_7=2048$ .

Explanation of Solution

Given: The sequence is $\frac{1}{2},-2,8,-32,\cdots$ .

Definition used:( Geometric Sequence).

A sequence $\left\{a_n\right\}$ is a geometric sequence and it can be written as $\left\{a,a\cdot r,a\cdot r^2,\cdots ,a\cdot r^{n-1},\cdots \right\}$ then the common ratio is r and the nth term is $a_n=a_{n-1}\cdot r\text{ for all }n\ge 2$ .

Calculation:

The first term is $a=\frac{1}{2}$ , the common ratio is $r=-4$ .

Substitute $a=\frac{1}{2}$ and $r=-4$ in $a_n=a\cdot r^{n-1}$ .

  $\begin{array}{l}{a}_7=\frac{1}{2}\cdot {\left(-4\right)}^{7-1}\\ =\frac{1}{2}\cdot {\left(-4\right)}^6\\ =\frac{4096}{2}\\ =2048\end{array}$

Thus, the seventh term of the sequence is $a_7=2048$ .

Therefore, the seventh term of the sequence $\frac{1}{2},-2,8,-32,\cdots$ is $a_7=2048$ .

(c)

To determine

The explicit rule for the nth term of the given sequence.

Answer to Problem 4RE

The explicit rule for the nth term of the sequence $\frac{1}{2},-2,8,-32,\cdots$ is $t_n=\left(\frac{1}{2}\right)\cdot {\left(-4\right)}^{n-1}$ .

Explanation of Solution

Given: The sequence is $\frac{1}{2},-2,8,-32,\cdots$ .

Definition used:( Geometric Sequence).

A sequence $\left\{a_n\right\}$ is a geometric sequence and it can be written as $\left\{a,a\cdot r,a\cdot r^2,\cdots ,a\cdot r^{n-1},\cdots \right\}$ then the common ratio is r and the nth term is $a_n=a_{n-1}\cdot r\text{ for all }n\ge 2$ .

Calculation:

Substitute $a=\frac{1}{2}\text{ and }r=-4$ in $t_n=a\cdot r^{n-1}$ as $t_n=\left(\frac{1}{2}\right)\cdot {\left(-4\right)}^{n-1}$ .

Therefore, the explicit rule for the nth term of the sequence $\frac{1}{2},-2,8,-32,\cdots$ is $t_n=\left(\frac{1}{2}\right)\cdot {\left(-4\right)}^{n-1}$ .