Chapter 5.8, Problem 35CCR

(a)

To determine

To find:Thelist of first ten terms in the numerical pattern.

Answer to Problem 35CCR

The first ten terms in the numerical pattern A is $0,4,8,12,16,20,24,28,32,36$ and in pattern B is $0,8,16,24,32,40,48,56,64,72$ .

Explanation of Solution

Given information:

Pattern A Use the standard number 0 and add 4.

Pattern B Use the standard number 0 and add 8

Calculation:

Pattern A $=0,4,8,12,16,20,24,28,32,36$

Pattern B $=0,8,16,24,32,40,48,56,64,72$

Therefore, the first ten terms in the numerical pattern A is $0,4,8,12,16,20,24,28,32,36$ and in pattern B is $0,8,16,24,32,40,48,56,64,72$ .

In term Pattern B, each term is twice as much as the term in Pattern A

Therefore, each term in Pattern B is twice as much as the term in Pattern A.

(b)

To determine

To explain:The relation between these two patterns.

Answer to Problem 35CCR

The number will be twice for any integer.

Explanation of Solution

Given information:

Pattern A Use the standard number 0 and add 4.

Pattern B Use the standard number 0 and add 8

Calculation:

For Pattern A the nth term will be worth,

  $\begin{array}{l}4n:n=0:4\cdot 0,\\ n=1:4\cdot 1=4\end{array}$ and so on

For Pattern B, that will,

  $\begin{array}{l}8n:n=0:8\cdot 0,\\ n=1:8\cdot 1\end{array}$ and so on

So for any n,

  $\to \frac{8n}{4n}=2$ .

Therefore, the number will be twice for any integer.