Chapter 12, Problem 1PRE

For Exercises 1−10, determine if the nth term of the sequence defines an arithmetic sequence, a geometric sequence, or neither. If the sequence is arithmetic, find the common difference d. If the sequence is geometric, find the common ratio r.

1. $a_n=\frac{2n}{3}+\frac{1}{4}$

To determine

To find the given equation sequence is arithmatric expression, a geomatric sequnce, or Neither.

Answer to Problem 1PRE

Given equation sequence is arithmatric expression with difference $d=\frac{2}{3}$

Explanation of Solution

Given:

Given equation is $a_n=\frac{2n}{3}+\frac{1}{4}$

Calculation:

Equation is $a_n=\frac{2n}{3}+\frac{1}{4}$

  $a_1=\frac{2}{3}+\frac{1}{4}$  [Using $n=1$]

  $a_2=\frac{4}{3}+\frac{1}{4}$  [Using $n=2$]

  $a_3=\frac{6}{3}+\frac{1}{4}$  [Using $n=3$]

  $a_4=\frac{8}{3}+\frac{1}{4}$  [Using $n=4$]

  $a_5=\frac{10}{3}+\frac{1}{4}$  [Using $n=5$]

So,now calculating the difference in the equation:

  $a_2-a_1=\frac{4}{3}+\frac{1}{4}-\frac{2}{3}-\frac{1}{4}$

  $a_2-a_1=\frac{2}{3}$

  $a_3-a_2=\frac{6}{3}+\frac{1}{4}-\frac{4}{3}-\frac{1}{4}$

  $a_3-a_2=\frac{2}{3}$

  $a_4-a_3=\frac{8}{3}+\frac{1}{4}-\frac{6}{3}-\frac{1}{4}$

  $a_4-a_3=\frac{2}{3}$

Now as $(a_2-a_1)=(a_3-a_2)=(a_4-a_3)=\frac{2}{3}$

Given equation sequence is arithmatric expression with difference $d=\frac{2}{3}$

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