Program for sum of arithmetic series
Last Updated :
16 Feb, 2023
A series with the same common difference is known as arithmetic series. The first term of series is a and the common difference is d. The series looks like a, a + d, a + 2d, a + 3d, . . . Task is to find the sum of the series.
Examples:
Input : a = 1
d = 2
n = 4
Output : 16
1 + 3 + 5 + 7 = 16
Input : a = 2.5
d = 1.5
n = 20
Output : 335
A simple solution to find the sum of arithmetic series.
C++
#include<bits/stdc++.h>
using namespace std;
float sumOfAP( float a, float d, int n)
{
float sum = 0;
for ( int i=0;i<n;i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
int main()
{
int n = 20;
float a = 2.5, d = 1.5;
cout<<sumOfAP(a, d, n);
return 0;
}
|
Java
class GFG{
static float sumOfAP( float a, float d,
int n)
{
float sum = 0 ;
for ( int i = 0 ; i < n; i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
public static void main(String args[])
{
int n = 20 ;
float a = 2 .5f, d = 1 .5f;
System.out.println(sumOfAP(a, d, n));
}
}
|
Python
def sumOfAP( a, d,n) :
sum = 0
i = 0
while i < n :
sum = sum + a
a = a + d
i = i + 1
return sum
n = 20
a = 2.5
d = 1.5
print (sumOfAP(a, d, n))
|
C#
using System;
class GFG {
static float sumOfAP( float a, float d,
int n)
{
float sum = 0;
for ( int i = 0; i < n; i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}
public static void Main()
{
int n = 20;
float a = 2.5f, d = 1.5f;
Console.Write(sumOfAP(a, d, n));
}
}
|
PHP
Javascript
Output:
335
Time Complexity: O(n)
Space complexity: O(1) because using constant space
Approach 2:
An Efficient solution to find the sum of arithmetic series is to use the below formula as follows:
Sum of arithmetic series
= ((n / 2) * (2 * a + (n - 1) * d))
Where
a - First term
d - Common difference
n - No of terms
Example:
C++
Java
Python3
C#
using System;
class GFG {
static float sumOfAP( float a,
float d,
float n)
{
float sum = (n / 2) *
(2 * a +
(n - 1) * d);
return sum;
}
static public void Main ()
{
float n = 20;
float a = 2.5f, d = 1.5f;
Console.WriteLine(sumOfAP(a, d, n));
}
}
|
PHP
<?php
function sumOfAP( $a , $d , $n )
{
$sum = ( $n / 2) * (2 * $a +
( $n - 1) * $d );
return $sum ;
}
$n = 20;
$a = 2.5; $d = 1.5;
echo (sumOfAP( $a , $d , $n ));
?>
|
Javascript
Time Complexity: O(1)
Space complexity: O(1) since using only constant variables
How does this formula work?
We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.
Let the formula be true for n = k-1.
Sum of first k - 1 elements of arithmetic series is
= (((k-1))/ 2) * (2 * a + (k - 2) * d))
We know k-th term of arithmetic series is
= a + (k - 1)*d
Sum of first k elements =
= Sum of (k-1) numbers + k-th element
= (((k-1)/2)*(2*a + (k-2)*d)) + (a + (k-1)*d)
= [((k-1)(2a + (k-2)d) + (2a + 2kd - 2d)]/2
= ((k / 2) * (2 * a + (k - 1) * d))
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