Chapter 12.1, Problem 2CFU

a.

To determine

To graph: The first five terms for the sequence $a_n=\frac{5-2n}{2}$ .

Explanation of Solution

Given information:

The sequence $a_n=\frac{5-2n}{2}$ .

Graph:

Consider the sequence $a_n=\frac{5-2n}{2}$ .

For first five terms substitute values of n in the above expression.

For first term substitute n as 1.

  $\begin{array}{c}{a}_1=\frac{5-2\left(1\right)}{2}\\ =\frac{3}{2}\end{array}$

For second term substitute n as 2.

  $\begin{array}{c}{a}_2=\frac{5-2\left(2\right)}{2}\\ =\frac{5-4}{2}\\ =\frac{1}{2}\end{array}$

For third term substitute n as 3.

  $\begin{array}{c}{a}_3=\frac{5-2\left(3\right)}{2}\\ =\frac{5-6}{2}\\ =\frac{-1}{2}\end{array}$

For fourth term substitute n as 4.

  $\begin{array}{c}{a}_4=\frac{5-2\left(4\right)}{2}\\ =\frac{5-8}{2}\\ =\frac{-3}{2}\end{array}$

For fifth term substitute n as 5.

  $\begin{array}{c}{a}_5=\frac{5-2\left(5\right)}{2}\\ =\frac{5-10}{2}\\ =\frac{-5}{2}\end{array}$

Denote the x-coordinate as n and y-coordinate as $a_n$ . Plot the values obtained above in a coordinate plane.

  

Interpretation:

The graph obtained by plotting the points for the first five terms of the sequence $a_n=\frac{5-2n}{2}$ is a straight line.

b.

To determine

To describe: The graph of first five terms for the sequence $a_n=\frac{5-2n}{2}$ .

Answer to Problem 2CFU

The graph is a straight line and points are plotted at equal distances from one other.

Explanation of Solution

Given information:

The sequence $a_n=\frac{5-2n}{2}$ .

Consider the sequence $a_n=\frac{5-2n}{2}$ .

For first five terms substitute values of n in the above expression.

For first term substitute n as 1.

  $\begin{array}{c}{a}_1=\frac{5-2\left(1\right)}{2}\\ =\frac{3}{2}\end{array}$

For second term substitute n as 2.

  $\begin{array}{c}{a}_2=\frac{5-2\left(2\right)}{2}\\ =\frac{5-4}{2}\\ =\frac{1}{2}\end{array}$

For third term substitute n as 3.

  $\begin{array}{c}{a}_3=\frac{5-2\left(3\right)}{2}\\ =\frac{5-6}{2}\\ =\frac{-1}{2}\end{array}$

For fourth term substitute n as 4.

  $\begin{array}{c}{a}_4=\frac{5-2\left(4\right)}{2}\\ =\frac{5-8}{2}\\ =\frac{-3}{2}\end{array}$

For fifth term substitute n as 5.

  $\begin{array}{c}{a}_5=\frac{5-2\left(5\right)}{2}\\ =\frac{5-10}{2}\\ =\frac{-5}{2}\end{array}$

Denote the x-coordinate as n and y-coordinate as $a_n$ . Plot the values obtained above in a coordinate plane.

  

The graph obtained by plotting the points for the first five terms of the sequence $a_n=\frac{5-2n}{2}$ is a straight line. All points are equally placed. The distance between each point is same. AS the value of n increase the value of $a_n=\frac{5-2n}{2}$ decreases.

c.

To determine

To calculate: The common differenceof the sequence $a_n=\frac{5-2n}{2}$ .

Answer to Problem 2CFU

The common differenceis $-1$ .Each term is 1 less than the preceding term. The magnitude of difference between two terms is 1.

Explanation of Solution

Given information:

The sequence $a_n=\frac{5-2n}{2}$ .

Calculation:

Consider the sequence $a_n=\frac{5-2n}{2}$ .

For first five terms substitute values of n in the above expression.

For first term substitute n as 1.

  $\begin{array}{c}{a}_1=\frac{5-2\left(1\right)}{2}\\ =\frac{3}{2}\end{array}$

For second term substitute n as 2.

  $\begin{array}{c}{a}_2=\frac{5-2\left(2\right)}{2}\\ =\frac{5-4}{2}\\ =\frac{1}{2}\end{array}$

For third term substitute n as 3.

  $\begin{array}{c}{a}_3=\frac{5-2\left(3\right)}{2}\\ =\frac{5-6}{2}\\ =\frac{-1}{2}\end{array}$

For fourth term substitute n as 4.

  $\begin{array}{c}{a}_4=\frac{5-2\left(4\right)}{2}\\ =\frac{5-8}{2}\\ =\frac{-3}{2}\end{array}$

For fifth term substitute n as 5.

  $\begin{array}{c}{a}_5=\frac{5-2\left(5\right)}{2}\\ =\frac{5-10}{2}\\ =\frac{-5}{2}\end{array}$

The sequence is $\frac{3}{2},\frac{1}{2},\frac{-1}{2},\frac{-3}{2},\frac{-5}{2}$ .

Recall that a sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

Compute the difference between the consecutive terms,

  $\begin{array}{c}{a}_2-a_1=\frac{1}{2}-\frac{3}{2}\\ =-1\end{array}$

And

  $\begin{array}{c}{a}_3-a_2=\frac{-1}{2}-\frac{-1}{2}\\ =-1\end{array}$

Therefore, common difference is $-1$ .

Each term is 1 less than the preceding term. The magnitude of difference between two terms is 1.

Therefore, it is an arithmetic sequence which is decreasing.