Chapter 8, Problem 20CR

To determine

To verify: the sequence is arithmetic or not.

Answer to Problem 20CR

The sequence $\frac{9}{9},\frac{8}{9},\frac{7}{9},\frac{6}{9},\frac{5}{9}\mathrm{....}$ is Arithmetic sequence having common difference of $-\frac{1}{9}$ .

Explanation of Solution

Given information:

The sequence which needs to be verified is as follows.

  $\frac{9}{9},\frac{8}{9},\frac{7}{9},\frac{6}{9},\frac{5}{9}\mathrm{....}$

Formula used:

Common difference of the arithmetic sequence is difference of its two consecutive terms and must be equal for all consecutive terms .

Calculation:

The sequence is as follows.

  $\frac{9}{9},\frac{8}{9},\frac{7}{9},\frac{6}{9},\frac{5}{9}\mathrm{....}$

First term of sequence is $a_1=\frac{9}{9}$ .

Second term of sequence is $a_2=\frac{8}{9}$ .

Third term of sequence is $a_3=\frac{7}{9}$ .

Fourth term of sequence is $a_4=\frac{6}{9}$ .

Fifth term of sequence is $a_5=\frac{5}{9}$ .

Calculating common difference

  $a_2-a_1=\frac{8}{9}-\frac{9}{9}=-\frac{1}{9}$ ,

  $a_3-a_2=\frac{7}{9}-\frac{8}{9}=-\frac{1}{9}$

  $a_4-a_3=\frac{6}{9}-\frac{7}{9}=-\frac{1}{9}$

  $a_5-a_4=\frac{5}{9}-\frac{6}{9}=-\frac{1}{9}$

As difference $a_2-a_1=a_3-a_2=a_4-a_3=a_5-a_4=-\frac{1}{9}$ , so the sequence is arithmetic with common difference of $-\frac{1}{9}$ .

Hence it can be concluded that the sequence $\frac{9}{9},\frac{8}{9},\frac{7}{9},\frac{6}{9},\frac{5}{9}\mathrm{....}$ is Arithmetic sequence having common difference of $-\frac{1}{9}$ .