Chapter 8.3, Problem 92E

To determine

To find whether the sequence associated with $\frac{5}{4}+\frac{7}{4}+\frac{9}{4}+\frac{11}{4}+\mathrm{...}$ is geometric or arithmetic and if so, then find its common ratio or common difference.

Answer to Problem 92E

The given sequence is an arithmetic sequence.

The common difference of the arithmetic sequence is $\frac{1}{2}$ .

Explanation of Solution

Given information:

An sequence associated with the seriesas $\frac{5}{4}+\frac{7}{4}+\frac{9}{4}+\frac{11}{4}+\mathrm{...}$ .

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

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

Common difference can be defined by d .

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

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Calculation:

Now, consider the given series.

  $\frac{5}{4}+\frac{7}{4}+\frac{9}{4}+\frac{11}{4}+\mathrm{...}$

The terms of the sequence can be found as

  $\begin{array}{l}{a}_1=\frac{5}{4}\\ a_2=\frac{7}{4}\\ a_3=\frac{9}{4}\\ a_4=\frac{11}{4}\\ \mathrm{...}\end{array}$

Now, the difference can be calculated as shown:

  $\begin{array}{l}{d}_1=a_2-a_1=\frac{7}{4}-\frac{5}{4}=\frac{1}{2}\\ d_2=a_3-a_2=\frac{9}{4}-\frac{7}{4}=\frac{1}{2}\\ d_3=a_4-a_3=\frac{11}{4}-\frac{9}{4}=\frac{1}{2}\end{array}$

Now, the differences for each two consecutive terms are as

  $d_1=\frac{1}{2}=d_2=d_3=\mathrm{...}$

Therefore, differences are common for each two consecutive terms of the sequence.

So, it is an arithmetic sequence.

And, the common difference of the arithmetic sequence is $d=\frac{1}{2}$ .