An ordered list of numbers is called sequences. Each number is the sequence is called a term. The sequence usually has patterns that allow us to predict what the next term of the sequence will be. An arithmetic series is the sum of sequence in which each term is computed from the previous one by adding and subtracting a constant. Or we can say that an arithmetic progression can be defined as a sequence of numbers in which for every pair of consecutive terms, the second number is found by adding a constant number to the previous one.
In Arithmetic Series/Progression we come across three terms which are:
- Common difference(d)
- nth term(an)
- Sum of the first n terms(Sn)
Here, all the above three terms represent the property of the arithmetic progression.
For finding the common difference of the arithmetic progression following procedure is followed:
d = a2 – a1 = a3 – a2 = a4 – a3….. = an-an-1
Where, a1, a2, a3….an are the terms of the series and “d” is the common difference which can be positive, negative or zero.
Also, an arithmetic progression can also be written in the form of common difference as shown below:
a, a + d, a + 2d, a + 3d, a + 4d, ………. , a + (n – 1)d
Where, “a” is the first term of series.
There are two major formulas while reading arithmetic progression:
1) The nth term of arithmetic series
2) Sum of first n terms
The nth term of Arithmetic Series
The formula for nth term is as:
an = a + (n−1)d
where, a=first term
d = common difference
n = number of terms
an = nth term
Sum of first n terms
The sum of first “n” terms of the series can be easily found is we know the first term of the series and total terms. The formula for finding the sum of first “n” terms is:
Sn = n/2 [2a + (n−1)d]
Where a = first term
d = common difference
n = number of terms.
Arithmetic Series Sigma Notation
Sigma notation looks like the below:
Here in the above expression the ‘i’ describes the initial value. The ‘f’ describes the final value and the expression refers for the function and the ‘E’ symbol is the Greek symbol called sigma.
Here the value of n starts with ‘1’ and ends at ’10’. When we start putting the value of n we get the arithmetic series just like below:
10 + 13 + 16 + 19……+ 37
Now for sigma notation, there is the formula used to find the sum of arithmetic series given above
Sn = n/2(a1 + an)
Here ‘n’ is the number of terms in the series and a1 and an is the first and last term of the series respectively.
For the above example we get the following values:
n = 10
a1 = 10
an = 37
Putting the value in the above equation
Sn = 10(10+37)/2 = 235
Σ10n=1 (3n+7)= 235
Arithmetic Series Sum Expression
When we get the arithmetic series expression like below:
1 + 5 + 9 +……+ 45
Now here we know that the expression for summation is as follows:
Sn = n/2(a1 + an)
Here we know the value
a1 = 1, an = 45, d = (5-1) = 4, but we don’t know the value of ‘n’ so we’ll find it using the below method:
n = (an – a1)/d = (45 – 1)/4 = 11
Therefore, the value of the n is 11.
Counting the sum of the series goes as Sn = n/2(a1 + a
Sn = 11(1 + 45)/2 = 253
Arithmetic Series Recursive Formula
Recursive Formula gives to two information:
1) The first term of the sequence
2) The pattern rule to find any term from the term that comes before it
Suppose, we have the series 3, 5, 7….. then here the first term of the series is a1 = 3
Now, from above the series we see that the formula for an Will be as below:
If a1 = 3 than an = a(n-1) + 2
Therefore, we have to add ‘2’ to the previous term to get to next term of the series.
Hence, finding the rest of the term below:
a1 = 3, a2 = a1+2 = 3 + 2 = 5, a3 = a2 + 2 = 5 + 2 = 7, a4 = a3 + 2 = 7 + 2 = 9, a5 = a4 + 2 = 11… and so on.
Proof of finite arithmetic series formula
As we know that Sn = n/2(a1 + an)
As we know at Sn is the sum of the arithmetic series of ‘n’ terms, therefore, we can display it like below:
Sn = 1 + 2 + 3….+ (n-1) + n
Sn = n + (n-1) +…+ 3 + 2 + 1
Adding both the Sn series we get the following result:
2Sn = n+1 + 2+(n-1) + 3+(n-2)…..
2Sn = n(n+1)
Sn = n(n+1)/2
In the above formula, we find that sum of every term on the left-side turn out to be ‘n+1’ for ‘n’ times.
Hence, we can prove that Sn = n(n+1)/2
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