Chapter 11.4, Problem 27WE

To determine

To calculate:

The series of positive three digit integers which must be divisible by $5$

Answer to Problem 27WE

The series of positive three digit integers which must be divisible by $5$ is ${\displaystyle \sum _{n=20}^{199}5n}$ .

Explanation of Solution

Given information:

Series divisible by $5$

Calculation:

The three digit integer starts from $100$ and ends on $995$ . Therefore for the given problem’s condition:

The first term: $100$

The last term: $995$

The series of the numbers three digit integer divisible by $5$ so, need to use the formula $5n$ to calculate the lower and upper limit as:

  $\begin{array}{l}5n=100\\ n=20\end{array}$

And

  $\begin{array}{l}5n=995\\ n=199\end{array}$

Therefore, lower and upper limit is $20$ and $199$ . Now, the series in sigma notation is expressed as:

  ${\displaystyle \sum _{n=20}^{199}5n}$