Chapter 8, Problem 13CT

To determine

To find: the series is geometric or arithmetic and the sum of series.

Answer to Problem 13CT

The series $5-2+\frac{4}{5}-\frac{8}{25}+\frac{16}{125}-\mathrm{...}$ is geometric series having common ratio $\underset{\_}{r=-\frac{2}{5}}$ and sum of the series is $\underset{\_}{\frac{25}{7}}$ .

Explanation of Solution

Given information:

The sequence which needs to be verified is as follows.

  $5-2+\frac{4}{5}-\frac{8}{25}+\frac{16}{125}-\mathrm{...}$

Formula used:

1. Common difference of the arithmetic sequence is difference of its two consecutive terms and must be equal for all consecutive terms.
2. Common ratio of the geometric sequence is ratio of its two Consecutive terms and must be equal for all consecutive terms.
3. Sum of infinite geometric series $S_n=\frac{a_1}{1-r}$ where $a_1$ is first term and $r$ is common ratio.

Calculation:

The series is as follows.

  $5-2+\frac{4}{5}-\frac{8}{25}+\frac{16}{125}-\mathrm{...}$

First term of series is $a_1=5$ .

Second term of series is $a_2=-2$ .

Third term of series is $a_3=\frac{4}{5}$ .

Fourth term of series is $a_4=-\frac{8}{25}$ .

Fifth term of series is $a_5=\frac{16}{125}$ .

Calculating ratio of consecutive terms

  $\frac{a_2}{a_1}=\frac{-2}{5}=-\frac{2}{5}$ ,

  $\frac{a_3}{a_2}=\frac{\left(\frac{4}{5}\right)}{-2}=-\frac{4}{2\times 5}=-\frac{2}{5}$

  $\frac{a_4}{a_3}=\frac{\left(-\frac{8}{25}\right)}{\left(\frac{4}{5}\right)}=-\frac{8}{25}\times \frac{5}{4}=-\frac{2}{5}$

  $\frac{a_5}{a_4}=\frac{\left(\frac{16}{125}\right)}{\left(-\frac{8}{25}\right)}=\frac{16}{125}\times \left(-\frac{25}{8}\right)=-\frac{2}{5}$

As ratio $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\frac{a_5}{a_4}=-\frac{2}{5}$ , so the sequence is geometric with common ratio that is $r=-\frac{2}{5}$ .

Calculating sum of the series

As $5-2+\frac{4}{5}-\frac{8}{25}+\frac{16}{125}-\mathrm{...}$ is infinite series, calculate the sum by using formula of infinite geometric series.

Substitute $a_1=5$ and $r=-\frac{2}{5}$ in formula $S_n=\frac{a_1}{1-r}$ .

  $S_n=\frac{5}{1-\left(-\frac{2}{5}\right)}$

  $=\frac{5}{1+\frac{2}{5}}$

  $=\frac{5}{\frac{5+2}{5}}$

  $=\frac{5\times 5}{7}$

  $=\frac{25}{7}$

Hence it can be concluded that the series $5-2+\frac{4}{5}-\frac{8}{25}+\frac{16}{125}-\mathrm{...}$ is geometric series having common ratio $\underset{\_}{r=-\frac{2}{5}}$ and sum of the series is $\underset{\_}{\frac{25}{7}}$ .