Chapter 8.2, Problem 63E

To determine

To find:The sum of the integers from $-100$ to 30.

Answer to Problem 63E

The sum of the integers from $-100$ to 30is $-4585$ .

Explanation of Solution

Given information:

Integers from $-100$ to 30.

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Sum of an arithmetic finite sequence has the form

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

Here, n is number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last term of the sequence,

Calculation:

Consider the given sequence.

Integers from $-100$ to 30.

The sequence will be

  $-100,-99,-98,-97,\mathrm{...},0,1,2,3,\mathrm{...},30$

Now, first term of the finite sequence is $a_1=-100$ .

Last term of the sequence is $a_n=30$ .

Common difference is

  $\begin{array}{c}d=a_2-a_1\\ =-99-\left(-100\right)\\ =-99+100\\ =1\end{array}$

Number of term of the sequence can be found as

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ 30=-100+\left(n-1\right)\left(1\right)\\ 30=-100+n-1\\ n=30+101\\ n=131\end{array}$

So, the sum of the finite sequence is calculated as

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{131}=\frac{131}{2}\left(-100+30\right)\\ =131\times \left(-35\right)\\ =-4585\end{array}$

Thus, the sum of the integers from $-100$ to 30is $-4585$ .