Chapter 9, Problem 6PS

(a)

To determine

To write: The first eight terms of the related arithmetic sequence and the nth term of the sequence.

Answer to Problem 6PS

The first eight terms of the sequence are 3, 5, 7, 9, 11, 13, 15, 17 and the nth term is $2n+1$ .

Explanation of Solution

Given:

The sequence of perfect squares 1, 4, 9, 16, 25, 36, 49, 64, 81,....

Calculation:

Let the arithmetic sequence be represented by $\left(a_n\right)$ .

The related sequence is obtained by subtracting the consecutive terms of the given sequence of perfect squares, as shown below.

  $\begin{array}{c}{a}_1=4-1\\ =3\\ =2\left(1\right)+1\end{array}$

  $\begin{array}{c}{a}_2=9-4\\ =5\\ =2\left(2\right)+1\end{array}$

  $\begin{array}{c}{a}_3=16-9\\ =7\\ =2\left(3\right)+1\end{array}$

So, the nth term of the sequence can be given by $a_n=2n+1$ and the first eight terms of the sequence are 3, 5, 7, 9, 11, 13, 15, 17.

(b)

To determine

To describe: The method to find an arithmetic sequence that is related to the given sequence of perfect cubes.

Answer to Problem 6PS

The description is given below.

Explanation of Solution

Given:

The sequence of perfect cubes is 1, 8, 27, 64, 216, 343, 512, 729,....

Calculation:

In order to obtain an arithmetic sequence related to the sequence of perfect sequence, first find the difference between the consecutive terms of the given sequence of cubes.

  $\begin{array}{c}8-1=7\\ 27-8=19\\ 64-27=37\\ 125-64=61\end{array}$

  $\begin{array}{c}216-125=91\\ 343-216=127\\ 512-343=169\\ 729-512=217\end{array}$

The obtained sequence is 7, 19, 37, 61, 91, 127, 169, 217,... which is not arithmetic. Take the difference of the consecutive terms again to observe the obtained sequence.

  $\begin{array}{c}19-7=12\\ 37-19=18\\ 61-37=24\\ 91-61=30\end{array}$

  $\begin{array}{c}127-91=36\\ 169-127=42\\ 217-169=48\end{array}$

The obtained sequence is 12, 18, 24, 30, 36, 42, 48,..... The common difference between the consecutive terms is 6. So, the above sequence qualifies as an arithmetic sequence with first term 12 and common difference 12.

(c)

To determine

To write: The first seven terms of the related sequence in part (b) and the nth term of the sequence.

Answer to Problem 6PS

The first seven terms are 12, 18, 24, 30, 36, 42, 48 and the nth term is $a_n=6n+6$ .

Explanation of Solution

Calculation:

As calculated in part (b), the obtained related sequence is 12, 18, 24, 30, 36, 42, 48,.....

In order to determine the common difference, let the sequence be represented by $\left(a_n\right)$ . Then,

  $\begin{array}{c}{a}_1=12\\ =6\left(1\right)+6\end{array}$

  $\begin{array}{c}{a}_2=18\\ =6\left(2\right)+6\end{array}$

  $\begin{array}{c}{a}_3=24\\ =6\left(3\right)+6\end{array}$

So, the nth term is represented as $a_n=6n+6$ .

(d)

To determine

To describe: The method to find an arithmetic sequence that is related to the given sequence of perfect fourth powers.

Answer to Problem 6PS

The description is given below.

Explanation of Solution

Given:

The sequence of perfect fourth powers is 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561,....

Calculation:

In order to obtain an arithmetic sequence related to the sequence of perfect sequence, first find the difference between the consecutive terms of the given sequence of fourth powers.

  $\begin{array}{c}16-1=15\\ 81-16=65\\ 256-81=175\\ 625-256=369\end{array}$

  $\begin{array}{c}1296-625=671\\ 2401-1296=1105\\ 4096-2401=1695\\ 6561-4096=2465\end{array}$

The obtained sequence is 15, 65, 175, 369, 671, 1105, 1695, 2465,... which is not arithmetic. Take the difference of the consecutive terms again to observe the obtained sequence.

  $\begin{array}{c}65-15=50\\ 175-65=110\\ 369-175=194\\ 671-369=302\end{array}$

  $\begin{array}{c}1105-671=434\\ 1695-1105=590\\ 2465-1695=770\end{array}$

The obtained sequence is 50, 110, 194, 302, 434, 590, 770,... which is still not arithmetic. Once again, take the difference of the consecutive terms to observe the obtained sequence.

  $\begin{array}{c}110-50=60\\ 194-110=84\\ 302-194=108\\ 434-302=132\end{array}$

  $\begin{array}{c}590-434=156\\ 770-590=180\end{array}$

The obtained sequence is 60, 84, 108, 132, 156, 180,..... The common difference between the consecutive terms is 24. So, the above sequence qualifies as an arithmetic sequence with first term 60 and common difference 24.

(e)

To determine

To write: The first six terms of the related sequence in part (d) and the nth term of the sequence.

Answer to Problem 6PS

The first six terms are 60, 84, 108, 132, 156, 180 and the nth term is $a_n=24n+36$ .

Explanation of Solution

Calculation:

As calculated in part (d), the obtained related sequence is 60, 84, 108, 132, 156, 180,.....

In order to determine the common difference, let the sequence be represented by $\left(a_n\right)$ . Then,

  $\begin{array}{c}{a}_1=60\\ =24\left(1\right)+36\end{array}$

  $\begin{array}{c}{a}_2=84\\ =24\left(2\right)+36\end{array}$

  $\begin{array}{c}{a}_3=108\\ =24\left(3\right)+36\end{array}$

So, the nth term is represented as $a_n=24n+36$ .