Chapter 9, Problem 5PS

a.

To determine

State the common difference.

Answer to Problem 5PS

  $d$ .

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

  $a,a+d,a+2d,a+3d,\mathrm{...}$

The general term $n^{th}$ of the sequence is given by

  $a_n=a+\left(n-1\right)d$

Add a constant $C$ to each term of the sequence.

So, the resulting sequence will be

  $a+C,a+d+C,a+2d+C,a+3d+C,\mathrm{...}$

Then the general term of the sequencing will be

  $a_n=a+\left(n-1\right)d+C$

Note that ,

  $a_n-a_{n-1}=\left(a+\left(n-1\right)d+C\right)-\left(a+\left(n-2\right)d+C\right)$

  $=d$

Difference between each two consecutive term is same,

Hence, the resulting sequence is also an arithmetic sequence and the coomon difference of arithmetic sequence is $d$ .

b.

To determine

State the common difference.

Answer to Problem 5PS

  $Cd$ .

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

  $a,a+d,a+2d,a+3d,\mathrm{...}$

The general term $n^{th}$ of the sequence is given by

  $a_n=a+\left(n-1\right)d$

Multiply each term of the sequence by $C$ .

So, the resulting sequence will be

  $aC,\left(a+d\right)C,\left(a+2d\right)C,\left(a+3d\right)C,\mathrm{...}$

Then the general term of the sequencing will be

  $a_n=\left(a+\left(n-1\right)d\right)C$

Note that ,

  $a_n-a_{n-1}=\left(a+\left(n-1\right)d\right)C-\left(a+\left(n-2\right)d\right)C$

  $=Cd$

Difference between each two consecutive term is same,

Hence, the resulting sequence is also an arithmetic sequence and the coomon difference of arithmetic sequence is $Cd$ .

c.

To determine

Find if each operation results in an arithmetic sequence.

Answer to Problem 5PS

The sequence is not an arithmetic sequence.

Explanation of Solution

Calculation:

Consider the arithmetic sequence with first term $a$ and common difference $d$ .

  $a,a+d,a+2d,a+3d,\mathrm{...}$

The general term $n^{th}$ of the sequence is given by

  $a_n=a+\left(n-1\right)d$

Square each term of the sequence.

So, the resulting sequence will be

  $a^2,{\left(a+d\right)}^2,{\left(a+2d\right)}^2,{\left(a+3d\right)}^2,\mathrm{...}$

Then the general term of the sequencing will be

  $a_n={\left(a+\left(n-1\right)d\right)}^2$

Note that ,

  $\begin{array}{l}{a}_n-a_{n-1}={\left(a+\left(n-1\right)d\right)}^2-{\left(a+\left(n-2\right)d\right)}^2\\ =\left(a^2+{\left(n-1\right)}^2{d}^2+2a\text{}\text{}\text{}\text{}\text{}\left(n-1\right)d-(a^2+{\left(n-2\right)}^2{d}^2+2a\text{}\text{}\text{}\text{}\text{}\left(n-2\right)d)\right)\\ =\left(-3+2n\right)d^2+2ad\end{array}$

So, the difference depends upon the value $n$ .

Difference between each two consecutive terms is not same.

Hence, the resulting sequence is not arithmetic sequence.