Chapter 10.1, Problem 39E

(a)

To determine

To find: The geometric series in the form ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ that will converge to 5 and a = 2,

Answer to Problem 39E

The geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(\frac{3}{5}\right)}^{n-1}}$ .

Explanation of Solution

Given:

The value of $a=2$ and the series converges to 5.

Concept used:

The mathematical expression of infinite geometric series in general is written as $a+ar+a{r}^2+a{r}^3+\cdots$ where the first term of the series is $a$ and the common ratio between two consecutive terms is $r$ .

The convergence test of series states that the geometric series converges if $\left|r\right|<1$ otherwise the series diverges where $r$ is the constant ratio.

Furthermore, the convergence test of geometric series also ensures that the convergent infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ converges to $\frac{a}{1-r}$ .

Calculation:

The infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ that converges to $S=\frac{a}{1-r}$ where $\left|r\right|<1$ .

Substitute 2 for a and 5 for S in the equation $S=\frac{a}{1-r}$ to obtain the value of $r$ .

  $\begin{array}{c}5=\frac{2}{1-r}\\ 5\left(1-r\right)=2\\ 5-5r=2\\ 5r=5-2\end{array}$

Further simplify the above equation.

  $\begin{array}{c}5r=3\\ r=\frac{3}{5}\end{array}$

Therefore, the value of $r=\frac{3}{5}$ .

Substitute 2 for a and $\frac{3}{5}$ for r in the series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ to obtain the requires series.

  ${\displaystyle \sum _{n=1}^{\infty }2{\left(\frac{3}{5}\right)}^{n-1}}$

Here, the initial term is 2 and each sequential term is multiplied by $\frac{3}{5}$ and the last term is the $n^{th}$ term that is equal to $2{\left(\frac{3}{5}\right)}^n$ .

Therefore, the geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(\frac{3}{5}\right)}^{n-1}}$ .

Conclusion:

Thus, the geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(\frac{3}{5}\right)}^{n-1}}$ .

(b)

To determine

To find: The geometric series in the form ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ that will converge to 5 and $a=\frac{13}{2}$ .

Answer to Problem 39E

The geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(-\frac{3}{10}\right)}^{n-1}}$ .

Explanation of Solution

Given:

The value of $a=\frac{13}{2}$ and the series converges to 5.

Concept used:

The mathematical expression of infinite geometric series in general is written as $a+ar+a{r}^2+a{r}^3+\cdots$ where the first term of the series is $a$ and the common ratio between two consecutive terms is $r$ .

The convergence test of series states that the geometric series converges if $\left|r\right|<1$ otherwise the series diverges where $r$ is the constant ratio.

Furthermore, the convergence test of geometric series also ensures that the convergent infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ converges to $\frac{a}{1-r}$ .

Calculation:

The infinite geometric series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ that converges to $S=\frac{a}{1-r}$ where $\left|r\right|<1$ .

Substitute $\frac{13}{2}$ for a and 5 for S in the equation $S=\frac{a}{1-r}$ to obtain the value of $r$ .

  $\begin{array}{c}5=\frac{\frac{13}{2}}{1-r}\\ 10\left(1-r\right)=13\\ 10-10r=13\\ 10r=10-13\end{array}$

Further simplify the above equation.

  $\begin{array}{c}10r=-3\\ r=-\frac{3}{10}\end{array}$

Therefore, the value of $r=-\frac{3}{10}$ .

Substitute 2 for a and $-\frac{3}{10}$ for r in the series ${\displaystyle \sum _{n=1}^{\infty }a}{r}^{n-1}$ to obtain the requires series.

  ${\displaystyle \sum _{n=1}^{\infty }2{\left(-\frac{3}{10}\right)}^{n-1}}$

Here, the initial term is 2 and each sequential term is multiplied by $-\frac{3}{10}$ and the last term is the $n^{th}$ term that is equal to $2{\left(-\frac{3}{10}\right)}^n$ .

Therefore, the geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(-\frac{3}{10}\right)}^{n-1}}$ .

Conclusion:

Thus, the geometric series is ${\displaystyle \sum _{n=1}^{\infty }2{\left(-\frac{3}{10}\right)}^{n-1}}$ .