a . 1 + 3 + 5 + 7 + 9 + ... + 99 = 2500
b. 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50
c. -100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50 = -3775
<h3>Further explanation</h3>
Let us learn about Arithmetic Progression.
Arithmetic Progression is a sequence of numbers in which each of adjacent numbers have a constant difference.
<em>Tn = n-th term of the sequence</em>
<em>Sn = sum of the first n numbers of the sequence</em>
<em>a = the initial term of the sequence</em>
<em>d = common difference between adjacent numbers</em>
Let us now tackle the problem!
<h2>Question a :</h2>
1 + 3 + 5 + 7 + 9 + ... + 99
<em>initial term = a = 1</em>
<em>common difference = d = ( 3 - 1 ) = 2</em>
Firstly , we will find how many numbers ( n ) in this series.
At last , we could find the sum of the numbers in the series using the above formula.
<h2>Question b :</h2>
In this question let us find the series of even numbers first , such as :
2 + 4 + 6 + 8 + ... + 100
<em>initial term = a = 2</em>
<em>common difference = d = ( 4 - 2 ) = 2</em>
<em />
Firstly , we will find how many numbers ( n ) in this series.
We could find the sum of the numbers in the series using the above formula.
At last , we could find the result of the series.
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100
= ( 1 + 3 + 5 + 7 + ... + 99 ) - ( 2 + 4 + 6 + 8 + ... + 100 )
= 2500 - 2550
= -50
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + ... - 100 = -50
<h2>Question c :</h2>
-100 - 99 - 98 - .... -2 - 1 - 0 + 1 + 2 + ... + 48 + 49 + 50
<em>initial term = a = -100</em>
<em>common difference = d = ( -99 - (-100) ) = 1</em>
<em />
Firstly , we will find how many numbers ( n ) in this series.
We could find the sum of the numbers in the series using the above formula.
<h3>Learn more</h3>
<h3>Answer details</h3>
Grade: Middle School
Subject: Mathematics
Chapter: Arithmetic and Geometric Series
Keywords: Arithmetic , Geometric , Series , Sequence , Difference , Term