Chapter 8.2, Problem 60E

To find The sum of the integer from −10 to 50.

To determine

The sum of the integerfrom −10 to 50 is $1220$ .

Given information:

Integers from −10 to 50.

Definition used:

An arithmetic sequence of n terms has the form $a_1,a_2,a_3,\cdots ,a_n$ and should be have a common difference between between each two consecutive terms which is given by,

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

The common difference can be defined by d is $a_2-a_1=a_3-a_2=a_4-a_3=\dots =a_n-a_{n-1}=d$

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 the number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last tems of sequence.

Calculation:

Find the sum of integers from −10 to 50.

The sequence will of the form $-10,-9,-8,\mathrm{..},0,1,2,3,\mathrm{..},50$ .

Compute the common difference as follows,

  $\begin{array}{c}d=a_2-a_1\\ =-9-\left(-10\right)\\ =-9+10\\ =1\end{array}$

So, the number of terms of the sequence will be,

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ 50=-10+\left(n-1\right)\left(1\right)\\ 50=-10+n-1\\ n=61\end{array}$

Therefore, the sum of the finite sequence is calculated as follows,

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ =\frac{61}{2}\left(-10+50\right)\\ =61\cdot \left(20\right)\\ =1220\end{array}$

Therefore, the sum of the integers from −10 or 50 is $1220$ .

Answer to Problem 60E

The sum of the integerfrom −10 to 50 is $1220$ .

Explanation of Solution

Given information:

Integers from −10 to 50.

Definition used:

An arithmetic sequence of n terms has the form $a_1,a_2,a_3,\cdots ,a_n$ and should be have a common difference between between each two consecutive terms which is given by,

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

The common difference can be defined by d is $a_2-a_1=a_3-a_2=a_4-a_3=\dots =a_n-a_{n-1}=d$

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 the number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last tems of sequence.

Calculation:

Find the sum of integers from −10 to 50.

The sequence will of the form $-10,-9,-8,\mathrm{..},0,1,2,3,\mathrm{..},50$ .

Compute the common difference as follows,

  $\begin{array}{c}d=a_2-a_1\\ =-9-\left(-10\right)\\ =-9+10\\ =1\end{array}$

So, the number of terms of the sequence will be,

  $\begin{array}{l}{a}_n=a_1+\left(n-1\right)d\\ 50=-10+\left(n-1\right)\left(1\right)\\ 50=-10+n-1\\ n=61\end{array}$

Therefore, the sum of the finite sequence is calculated as follows,

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ =\frac{61}{2}\left(-10+50\right)\\ =61\cdot \left(20\right)\\ =1220\end{array}$

Therefore, the sum of the integers from −10 or 50 is $1220$ .