
Concept explainers
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 a1,a2,a3,...,an and should have a common difference between each two consecutive terms.
That is
a2−a1=a3−a2=a4−a3=...=an−an−1
Common difference can be defined by d .
a2−a1=a3−a2=a4−a3=...=an−an−1=d
nth term of the arithmetic sequence has the form
an=a1+(n−1)d
Where a1 is the first term of the sequence, and d is the common difference.
Sum of an arithmetic finite sequence has the form
Sn=n2(a1+an)
Here, n is number of terms, a1 is the first term of the sequence, and an 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 a1=1 .
Last term of the sequence is an=100 .
Common difference is
d=a2−a1=2−1=1
Number of term of the sequence can be found as
an=a1+(n−1)d100=1+(n−1)(1)100=1+n−1n=100
So, the sum of the finite sequence is calculated as
Sn=n2(a1+an)S100=1002(1+100)=50×(101)=5050
Thus, the sum of first 100 positive integersis 5050 .
Chapter 8 Solutions
Precalculus with Limits: A Graphing Approach
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