Chapter 10, Problem 6CA

To determine

To Describe: The pattern, write first five terms and a rule for the $nth$ term.

Answer to Problem 6CA

A recursive rule for the sequence is $a_n=\frac{1}{n}{a}_1$

The $nth$ term is found by the rule is $a_n=-\frac{1}{2n}$ .

The next five terms of the sequence are $a_5=-\frac{1}{10}$

  $a_6=-\frac{1}{12}$

  $a_7=-\frac{1}{14}$

  $a_8=-\frac{1}{16}$

  $a_9=-\frac{1}{18}$

Explanation of Solution

Given:

  $\frac{-1}{2},\frac{-1}{4},\frac{-1}{6},\frac{-1}{8}\mathrm{...........}$

Calculation:

The terms have neither common ratio nor common difference.

Denote the first term by $a_1$ . Note that

  $\begin{array}{l}{a}_2=\frac{-1}{4}=\frac{1}{2}.a_1\\ a_3=\frac{-1}{6}=\frac{1}{3}.a_1\\ a_4=\frac{-1}{8}=\frac{1}{4}.a_1\\ \end{array}$

Thus, a recursive rule for the sequence is

  $a_1=\frac{-1}{2},$

  $a_n=\frac{1}{n}{a}_1$

The next five terms of the sequence are

  $\begin{array}{l}{a}_5=\frac{1}{5}{a}_1\\ a_5=\frac{1}{5}\left(-\frac{1}{2}\right)\end{array}$

  $a_5=-\frac{1}{10}$

  $\begin{array}{l}{a}_6=\frac{1}{6}{a}_1\\ a_6=\frac{1}{6}\left(-\frac{1}{2}\right)\\ a_6=-\frac{1}{12}\end{array}$

  $\begin{array}{l}{a}_7=\frac{1}{7}{a}_1\\ a_7=\frac{1}{7}\left(-\frac{1}{2}\right)\\ a_7=-\frac{1}{14}\end{array}$

  $\begin{array}{l}{a}_8=\frac{1}{8}{a}_1\\ a_8=\frac{1}{8}\left(-\frac{1}{2}\right)\\ a_8=-\frac{1}{16}\end{array}$

  $\begin{array}{l}{a}_9=\frac{1}{9}{a}_1\\ a_9=\frac{1}{9}\left(-\frac{1}{2}\right)\\ a_9=-\frac{1}{18}\end{array}$

The $nth$ term is found by the rule

  $a_n=\frac{1}{n}{a}_1=-\frac{1}{2n}$

Conclusion:

Therefore, a recursive rule for the sequence is $a_n=\frac{1}{n}{a}_1$

The $nth$ term is found by the rule is $a_n=-\frac{1}{2n}$ .

The next five terms of the sequence are $a_5=-\frac{1}{10}$

  $a_6=-\frac{1}{12}$

  $a_7=-\frac{1}{14}$

  $a_8=-\frac{1}{16}$

  $a_9=-\frac{1}{18}$