Chapter 7.3, Problem 8MRPS

Solution

To calculate : The arithmetic and geometric series such that both have five terms and the same sum.

The arithmetic and geometric series such that both have five terms and the same sum are $a_n=2+5.2\left(n-1\right)$ and $a_n=2{\left(2\right)}^{n-1}$ .

Given information :

The series has five terms and the same sum.

Formula used :

The $n^{th}$ term of an arithmetic series with first term as $a_1$ and common difference $d$ is given by $a_n=a_1+\left(n-1\right)d$ .

The $n^{th}$ term of a geometric series with first term as $a_1$ and common ratio $r$ is given by $a_n=a_1{r}^{n-1}$ .

Calculation :

As the series have five terms, so $n=2$ .

Assume $a_1=2$ and equate the sums of arithmetic and geometric series.

  $\begin{array}{c}{\displaystyle \sum _{i=1}^5{a}_1+\left(n-1\right)d}={\displaystyle \sum _{i=1}^5{a}_1{r}^{n-1}}\\ 5\left(\frac{2+a_5}{2}\right)=2\left(\frac{1-r^5}{1-r}\right)\\ 5\left(\frac{2+a_1+4d}{2}\right)=2\left(\frac{1-r^5}{1-r}\right)\\ 5\left(\frac{2+2+4d}{2}\right)=2\left(\frac{1-r^5}{1-r}\right)\end{array}$

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

Now, assume $r=2$ to calculate the value of $d$ .

  $\begin{array}{c}d=\frac{1}{5}\left(\frac{1-2^5}{1-2}\right)-1\\ =\frac{1}{5}\left(\frac{1-32}{-1}\right)-1\\ =\frac{1}{5}\left(\frac{-31}{-1}\right)-1\\ =0.2\left(31\right)-1\end{array}$

  $\begin{array}{c}d=6.2-1\\ =5.2\end{array}$

Substitute the values $a_1=2$ and $d=5.2$ to formula $a_n=a_1+\left(n-1\right)d$ .

  $\begin{array}{c}{a}_n=2+\left(n-1\right)5.2\\ =2+5.2\left(n-1\right)\end{array}$

Substitute the values $a_1=2$ and $r=2$ to formula $a_n=a_1{r}^{n-1}$ .

  $a_n=2{\left(2\right)}^{n-1}$

Thus, the arithmetic and geometric series such that both have five terms and the same sum are $a_n=2+5.2\left(n-1\right)$ and $a_n=2{\left(2\right)}^{n-1}$ .