Chapter 8.2, Problem 76E

To determine

To find:Thepartial sum ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

Answer to Problem 76E

  ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}=218625$

Explanation of Solution

Given information:

A sequence is given as ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$ .

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 partial sum.

  ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}$

The above sum can be written as

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}=\left(1000-1\right)+\left(1000-2\right)+\left(1000-3\right)+\left(1000-4\right)+\mathrm{...}+\left(1000-250\right)\\ =999+998+997+996+\mathrm{...}+750\end{array}$

Now, first term of the finite sequence is $a_1=999$ .

The difference is

  $\begin{array}{c}{d}_1=a_2-a_1\\ =998-999\\ =-1\\ d_2=a_3-a_2\\ =997-998\\ =-1\\ d_3=a_4-a_3\\ =996-997\\ =-1\\ d_1=d_2=d_3=\mathrm{...}\end{array}$

It is an arithmetic sequence.

Last term is $a_{250}=750$

Number of term for the required sumis indicated as 250.

So, partial sum is calculated as

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{250}=\frac{250}{2}\left(999+750\right)\\ =125\times 1749\\ =218625\end{array}$

Thus, ${\displaystyle \sum _{n=1}^{250}\left(1000-n\right)}=218625$ .