Chapter 8.2, Problem 17E

To determine

To identify: Whether the sequence $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ is arithmetic or not.

Answer to Problem 17E

The sequence $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ is arithmetic because the difference between each two term is common.

Explanation of Solution

Given:

The sequence $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$

Concept used:

A sequence is arithmetic when the differences between consecutive terms are the same. So, the sequence

  $a_1,a_2,a_3,\mathrm{....},a_n,\mathrm{.....}$ is arithmetic when there is a number $d$ such that

  $a_2-a_1=a_3-a_2=a_4-a_3=a_{n+1}-a_n=\mathrm{.....}=d$

The number $d$ is called the common difference of the arithmetic sequence.

Calculation:

First calculate the first five terms of the sequence.

Calculate the 1^st term $a_1$ by substituting $n=1$ in $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ .

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

Calculate the 2^nd term $a_2$ by substituting $n=2$ in $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ .

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

Calculate the 3^rd term $a_3$ by substituting $n=3$ in $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ .

  $\begin{array}{c}{a}_3={\left(-1\right)}^{2\left(3\right)}\left(\frac{1}{4}\right)\\ ={\left(-1\right)}^6\left(\frac{1}{4}\right)\\ =1\left(\frac{1}{4}\right)\\ =\frac{1}{4}\end{array}$

Calculate the 4^th term $a_4$ by substituting $n=4$ in $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ .

  $\begin{array}{c}{a}_4={\left(-1\right)}^{2\left(4\right)}\left(\frac{1}{4}\right)\\ ={\left(-1\right)}^8\left(\frac{1}{4}\right)\\ =1\left(\frac{1}{4}\right)\\ =\frac{1}{4}\end{array}$

Calculate the 5^th term $a_5$ by substituting $n=5$ in $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ .

  $\begin{array}{c}{a}_5={\left(-1\right)}^{2\left(5\right)}\left(\frac{1}{4}\right)\\ ={\left(-1\right)}^{10}\left(\frac{1}{4}\right)\\ =1\left(\frac{1}{4}\right)\\ =\frac{1}{4}\end{array}$

Hence, the sequence obtained from $a_n={\left(-1\right)}^{2n}\left(\frac{1}{4}\right)$ is $\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\mathrm{...}$ .

Calculate the difference between $a_2$ and $a_1$ .

  $\begin{array}{c}{d}_1=a_2-a_1\\ =\frac{1}{4}-\frac{1}{4}\\ =0\end{array}$

Since all the terms in the sequence being the same, the common difference $d_1=d_2=d_3=d_4=0$ .

Thus, from the above calculation the difference between each two term is common. So the sequence is arithmetic.