Chapter 8, Problem 3RE

(a)

To determine

The common difference for the given sequence.

Answer to Problem 3RE

The common difference of $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $d=\frac{3}{2}$ .

Explanation of Solution

Given: The sequence is $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ .

Definition used:( Arithmetic Sequence).

A sequence $\left\{a_n\right\}$ is an arithmetic sequence and it can be written as $\left\{a,a+d,a+2d+\cdots +a+\left(n-1\right)d+\cdots \right\}$ then the common difference is d and the nth term is $a_n=a_{n-1}+d\text{ for all }n\ge 2$ .

Calculation:

The terms are $t_1=-1,t_2=\frac{1}{2},t_3=2,t_4=\frac{7}{2},\cdots$

The common difference between the first two terms is,

  $\begin{array}{c}{t}_2-t_1=\frac{1}{2}-\left(-1\right)\\ =\frac{1}{2}+1\\ =\frac{3}{2}\end{array}$

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

  $\begin{array}{c}{t}_3-t_2=2-\frac{1}{2}\\ =\frac{4-1}{1}\\ =\frac{3}{2}\end{array}$ .

Similarly, the common difference for the consecutive terms of $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $d=\frac{3}{2}$

Therefore, the common difference of $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $d=\frac{3}{2}$ .

(b)

To determine

The tenth term for the given sequence.

Answer to Problem 3RE

The tenth term of the sequence $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $a_{10}=\frac{25}{2}$ .

Explanation of Solution

Given: The sequence is $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ .

Definition used:( Arithmetic Sequence).

A sequence $\left\{a_n\right\}$ is an arithmetic sequence and it can be written as $\left\{a,a+d,a+2d+\cdots +a+\left(n-1\right)d+\cdots \right\}$ then the common difference is d and the nth term is $a_n=a_{n-1}+d\text{ for all }n\ge 2$ .

Calculation:

The first term is $a=-1$ common difference is $d=\frac{3}{2}$ .

Substitute $a=-1$ and $d=\frac{3}{2}$ in $t_n=a+\left(n-1\right)d$ .

  $\begin{array}{c}{t}_{10}=\left(-1\right)+\left(10-1\right)\frac{3}{2}\\ =-1+\frac{30}{2}-\frac{3}{2}\\ =\frac{25}{2}\end{array}$

Therefore, the tenth term of the sequence $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $a_{10}=\frac{25}{2}$ .

(c)

To determine

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

Answer to Problem 3RE

The explicit rule for the nth term of the sequence $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $a_n=\frac{3n}{2}-\frac{5}{2}$ .

Explanation of Solution

Given: The sequence is $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ .

Definition used:( Arithmetic Sequence).

A sequence $\left\{a_n\right\}$ is an arithmetic sequence and it can be written as $\left\{a,a+d,a+2d+\cdots +a+\left(n-1\right)d+\cdots \right\}$ then the common difference is d and the nth term is $a_n=a_{n-1}+d\text{ for all }n\ge 2$ .

Calculation:

Substitute $a=-1\text{ and }d=\frac{3}{2}$ in $t_n=a+\left(n-1\right)d$ .

  $\begin{array}{c}{t}_n=\left(-1\right)+\left(n-1\right)\frac{3}{2}\\ =-1+\frac{3n}{2}-\frac{3}{2}\\ =\frac{3n}{2}-\frac{5}{2}\end{array}$

Therefore, the explicit rule for the nth term of the sequence $-1,\frac{1}{2},2,\frac{7}{2},\cdots$ is $a_n=\frac{3n}{2}-\frac{5}{2}$ .