Chapter 9, Problem 6PS

a.

To determine

Find the sequence.

Answer to Problem 6PS

  $3,5,7,9,11,13,15,17$

  $a_n=2n+1$

Explanation of Solution

Calculation:

Consider the sequence,

  $\text{1},\text{ 4},\text{ 9},\text{ 16},\text{ 25},\text{ 36},\text{ 49},\text{ 64},\text{ 81},\mathrm{....}$

Let us find the difference between the consecutive numbers, so,

  $\begin{array}{l}4-1=3\\ 9-4=5\\ 16-9=7\\ 25-16=9\end{array}$

The related sequence that is arithmetic sequence is,

  $3,5,7,9,11,13$

The above sequence is,

  $a_n=2n+1$

The first eight terms, given by sequence $a_n=2n+1$ is

  $\begin{array}{l}{a}_2=2\left(2\right)+1\\ a_2={\mathrm{5......2}}^{nd}term\\ a_3=2\left(3\right)+1\\ a_3={\mathrm{7......3}}^{rd}term\\ a_4=2\left(4\right)+1\\ a_4={\mathrm{9......4}}^{th}term\end{array}$

  $\begin{array}{l}{a}_5=2\left(5\right)+1\\ a_5={\mathrm{11......5}}^{th}term\\ a_6=2\left(6\right)+1\\ a_6={\mathrm{13......6}}^{th}term\\ a_7=2\left(7\right)+1\\ a_7={\mathrm{15......7}}^{th}term\\ a_8=2\left(8\right)+1\\ a_8={\mathrm{17......8}}^{th}term\end{array}$

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

b.

To determine

Find an arithmetic sequence.

Answer to Problem 6PS

  $12,18,24,30,36,42,48$

Explanation of Solution

Calculation:

The sequence of perfect cubes is,

  $\text{1},\text{ 8},\text{ 27},\text{ 64},\text{ 125},\text{ 216},\text{ 343},\text{ 512},\text{ 729},\mathrm{....}$

Let us find the difference between the consecutive numbers, so,

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

So on and so forth, which brings us to the following sequence:

  $7,19,37,61,91,127,169,217$

We need to again find the differences of the consecutive terms of the above sequence , that is,

  $12,18,24,30,36,42,48$

And the $n^{th}$ term is

  $a_n=6n+6$ .

c.

To determine

Write the first seven terms of the related arithmetic sequence in part $\left(b\right)$ and find the $n^{th}$ term of the sequence.

Answer to Problem 6PS

  $\text{12},\text{ 18},\text{ 24},\text{ 3}0,\text{ 36},\text{ 42},\text{ 48}$

  $a_n=6n+6$

Explanation of Solution

Calculation:

The sequence of the perfect cubes is

  $\text{1},\text{ 8},\text{ 27},\text{ 64},\text{ 125},\text{ 216},\text{ 343},\text{ 512},\text{ 729},\mathrm{....}$

As seen from (b), arithmetic sequence is

  $a_n=6n+6$

The first seven given by the sequence is,

  $\begin{array}{l}{a}_1=6\left(1\right)+6\\ a_1={\mathrm{12......1}}^{st}term\\ a_2=6\left(2\right)+6\\ a_2={\mathrm{18......2}}^{nd}term\\ a_3=6\left(3\right)+6\\ a_3={\mathrm{24......3}}^{rd}term\\ a_4=6\left(4\right)+6\\ a_4={\mathrm{30......4}}^{th}term\end{array}$

  $\begin{array}{l}{a}_5=6\left(5\right)+6\\ a_5={\mathrm{36......5}}^{th}term\\ a_6=6\left(6\right)+6\\ a_6={\mathrm{42......6}}^{th}term\\ a_7=6\left(7\right)+6\\ a_7={\mathrm{48......7}}^{th}ter\end{array}$

Hence, the first seven terms, given by

  $\text{12},\text{ 18},\text{ 24},\text{ 3}0,\text{ 36},\text{ 42},\text{ 48}$ , $a_n=6n+6$

d.

To determine

Find an arithmetic sequence that is related to the following sequence of perfect fourth powers.

Answer to Problem 6PS

  $\text{6}0,\text{ 84},\text{ 1}0\text{8},\text{ 132},\text{ 156},\text{ 18}0$

Explanation of Solution

Calculation:

Sequence of perfect fourth powers is

  $\text{1},\text{ 16},\text{ 81},\text{ 256},\text{ 625},\text{ 1296},\text{ 24}0\text{1},\text{ 4}0\text{96},\text{ 6561},\mathrm{....}$

Differences in consecutive terms of the sequence of fourth powers is,

  $\text{15},\text{ 65},\text{ 175},\text{ 369},\text{ 671},\text{ 11}0\text{5},\text{ 1695},\text{ 2645}$

Again taking differences in consecutive terms of the sequence above, we get

  $\text{5}0,\text{ 11}0,\text{ 194},\text{ 3}0\text{2},\text{ 434},\text{ 59}0,\text{ 77}0$

Again taking differences in consecutive terms of the sequence above, we get

  $\text{6}0,\text{ 84},\text{ 1}0\text{8},\text{ 132},\text{ 156},\text{ 18}0$

The $n^{th}$ term of the sequence is $a_n=24n+36$ .

Hence, the arithmetic sequence is,

  $\text{6}0,\text{ 84},\text{ 1}0\text{8},\text{ 132},\text{ 156},\text{ 18}0$

e.

To determine

Write the first six terms and find the $n^{th}$ term of the sequence.

Answer to Problem 6PS

  $\text{6}0,\text{ 84},\text{ 1}0\text{8},\text{ 132},\text{ 156},\text{ 18}0$

  $a_n=24n+36$

Explanation of Solution

Calculation:

The $n^{th}$ term of the sequence is $a_n=24n+36$ .

The first six terms of the sequence is given by

  $\begin{array}{l}{a}_1=24\left(1\right)+36\\ a_1={\mathrm{60......1}}^{st}term\\ a_2=24\left(2\right)+36\\ a_2={\mathrm{84......2}}^{nd}term\\ a_3=24\left(3\right)+36\\ a_3={\mathrm{108......3}}^{rd}term\\ a_4=24\left(4\right)+36\\ a_4={\mathrm{132......4}}^{th}term\\ a_5=24\left(5\right)+36\\ a_5={\mathrm{156......5}}^{th}term\\ a_6=24\left(6\right)+36\\ a_6={\mathrm{180......6}}^{th}term\end{array}$

Hence, the first six terms of the sequence is given by $\text{6}0,\text{ 84},\text{ 1}0\text{8},\text{ 132},\text{ 156},\text{ 18}0$ and

  $n^{th}$ term of the sequence is $a_n=24n+36$ .