Chapter 11.4, Problem 30WE

To determine

To calculate:

An infinite arithmetic series

Answer to Problem 30WE

An infinite arithmetic series is ${\displaystyle \sum _{n=0}^{10}50n+8}$ or $8+58+108+\mathrm{...}+508$ .

Explanation of Solution

Given information:

First term: $8$

Common ratio: $508$

  $n=11$

Formula used:

The $n^{th}$ term of arithmetic sequence is given by the following expression:

  $a_n=a_1+\left(n-1\right)d$

Where,

First term: $a_1$

Common difference: $d$

Calculation:

The common difference is calculated as below:

  $a_{11}=a_1+\left(n-1\right)d$

Put the given data:

  $\begin{array}{l}508=8+\left(11-1\right)d\\ 10d=500\\ d=50\end{array}$

  $\begin{array}{l}{a}_2=a_1+\left(n-1\right)d=8+\left(2-1\right)50=8+50=58\\ a_3=a_2+\left(n-1\right)d=58+\left(2-1\right)50=58+50=108\\ a_4=a_3+\left(n-1\right)d=108+\left(2-1\right)50=108+50=158\end{array}$

So, formula becomes: $50n+8$

Now, the lower limit is calculated by using the first term:

  $\begin{array}{l}50n+8=8\\ 50n=0\\ n=0\end{array}$

Now, the upper limit is calculated by using the last term:

  $\begin{array}{l}50n+8=508\\ 50n=500\\ n=10\end{array}$

Hence, an arithmetic series is expressed as:

  ${\displaystyle \sum _{n=0}^{10}50n+8}=8+58+108+\mathrm{...}+508$