Chapter 8.2, Problem 61E

To determine

To find:The sum of the first 100 positive integers.

Answer to Problem 61E

The sum of first 100 positive integersis $5050$ .

Explanation of Solution

Given information:

First 100 positive integers,

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.

First 100 positive integers.

The first positive integer is 1.

The first 100 positive integers are

1, 2, 3, 4, 5,…, 100

Now, first term of the finite sequence is $a_1=1$ .

Last term of the sequence is $a_n=100$ .

Common difference is

  $\begin{array}{c}d=a_2-a_1\\ =2-1\\ =1\end{array}$

Number of term of the sequence can be found as

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

So, the sum of the finite sequence is calculated as

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(a_1+a_n\right)\\ S_{100}=\frac{100}{2}\left(1+100\right)\\ =50\times \left(101\right)\\ =5050\end{array}$

Thus, the sum of first 100 positive integersis $5050$ .