Chapter 12.4, Problem 1CFU

(a)

To determine

To find: an example of an infinite geometric series in which $\left|r\right|>1$ .

Answer to Problem 1CFU

The example is $3+9+27+81+\mathrm{......}$

Explanation of Solution

Given:

  $\left|r\right|>1$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k$ .

Calculation:

Let’s consider the series as:

  $3+9+27+81+\mathrm{......}$

The above series is a geometric series with $r=3>1$ .

This is an example of infinite geometric series in which $\left|r\right|>1$ .

Hence, the example is $3+9+27+81+\mathrm{......}$

(b)

To determine

To find: the $25\text{th,}50\text{th and }100\text{th}$ term of the series.

Answer to Problem 1CFU

  $t_{25}=8.47\times {10}^{11},t_{50}=8.47\times {10}^{11}\text{ and }{t}_{100}=5.15\times {10}^{47}.$

Explanation of Solution

Given:

  $\begin{array}{l}{25}^{th},{50}^{th}{\text{ and 100}}^{th}.\\ \left|r\right|>1.\end{array}$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\end{array}$

Calculation:

The ${25}^{th},{50}^{th}{\text{ and 100}}^{th}$ terms of the series.

  $3+9+27+81+\mathrm{......}$

The above series is a geometric series.

Here $r=3$ and to find the $nth$ term of the series by using the below formula as:

  $\begin{array}{l}{t}_n=a{r}^{n-1}.\\ \Rightarrow 3{\left(3\right)}^{n-1}.\\ \Rightarrow 3^{1+n-1}.\\ \Rightarrow 3^n.\end{array}$

  ${25}^{th},{50}^{th}{\text{ and 100}}^{th}$ terms of the series can be find using formula:

  $\begin{array}{l}{t}_{25}=3^{25}\\ \Rightarrow 8.47\times {10}^{11}.\\ t_{50}=3^{50}\\ \Rightarrow 7.17\times {10}^{23}.\\ t_{100}=3^{100}\\ \Rightarrow 5.15\times {10}^{47}.\end{array}$

Hence, $t_{25}=8.47\times {10}^{11},t_{50}=8.47\times {10}^{11}\text{ and }{t}_{100}=5.15\times {10}^{47}.$

(c)

To determine

To find: the sum of the first $25,50\text{ and }100$ terms of series.

Answer to Problem 1CFU

  $s_{25}=1.26\times {10}^{12},s_{50}=1.075\times {10}^{24}\text{ and }{s}_{100}=7.725\times {10}^{47}$ .

Explanation of Solution

Given:

  $\left|r\right|>1.$

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Calculation:

The sum of the first $25,50\text{ and }100$ terms of the series:

  $3+9+27+81+\mathrm{......}$

Here $r=3$ :

Using the below formulas for finding the sum of series as:

  $\begin{array}{l}{s}_n=\frac{r\left(1-r^n\right)}{1-r}.\\ \Rightarrow \frac{3\left(1-3^n\right)}{1-3}.\\ \Rightarrow \frac{3}{2}\left(3^n-1\right).\end{array}$

Hence, the sum series is $s_{25}=1.26\times {10}^{12},s_{50}=1.075\times {10}^{24}\text{ and }{s}_{100}=7.725\times {10}^{47}$ .

(d)

To determine

To find: the type of infinite geometric series which does not converge.

Answer to Problem 1CFU

These type of geometric series do not converge because, if $\left|r\right|>1,$ then $r^n$ increases without limit as $n\to \infty$ .

Explanation of Solution

Given:

  $\left|r\right|>1$ .

Concept used:

If there is common ratio then the series is geometric series.

Common ratio is:

  $\begin{array}{l}r=\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=k.\\ t_n=a{r}^{n-1}.\\ s_n=\frac{r\left(1-r^n\right)}{1-r}.\end{array}$

Geometric series never converge:

  $\left|r\right|>1,$ implies $r^n\text{ and }n\to \infty$ .

Calculation:

The sum of the first $25,50\text{ and }100$ terms of the series:

  $3+9+27+81+\mathrm{......}$

Here $r=3$ :

Hence, these type of geometric series do not converge because, if $\left|r\right|>1,$ then $r^n$ increases without limit as $n\to \infty$ .