Chapter 11.5, Problem 24WE

To determine

To calculate:

The sum value for the following condition:

Positive three digit integers divisible by $6$

Answer to Problem 24WE

The sum value of series of positive three digit odd numbers is $37050$ .

Explanation of Solution

Given information:

Positive three digit integers divisible by $6$

Calculation:

The positive three digit integers start from $100$ and ends on $999$ . But need to define the series of integers divisible by $6$ :

The first three digit integer is $102$ and last the last three integer is $996$ between $100$ and $999$ .

So, the series is defined by taking common difference of $6$ as these numbers are divisible by $6$ : $102,108,114,\mathrm{...},996$

Consider,

  $\begin{array}{l}{a}_1=102\\ a_n=996\end{array}$

Here, each value exceeds by six, so the formula becomes:

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ a_n=102+\left(n-1\right)\left(6\right)\\ a_n=102+6n-6\\ a_n=96+6n\end{array}$

Now, calculate the lower and upper limit of the summation notation:

Lower limit:

  $\begin{array}{l}{a}_1=a_1+\left(n-1\right)d\\ 102=102+\left(n-1\right)\left(6\right)\\ \left(n-1\right)\left(6\right)=0\\ n-1=0\\ n=1\end{array}$

And upper limit is:

  $\begin{array}{l}{a}_1=a_1+\left(n-1\right)d\\ 996=102+\left(n-1\right)\left(6\right)\\ 996=96+6n\\ 6n=900\\ n=150\end{array}$

Hence, the summation notation of the series of three digit numbers divisible by six is ${\displaystyle \sum _{n=1}^{150}96+6n}$

Now, by using the summation calculator the sum value of series of positive three digit integers divisible by $6$ is ${\displaystyle \sum _{n=1}^{150}96+6n}=37050$ .