Program for sum of arithmetic series

• Difficulty Level : Easy
• Last Updated : 01 Apr, 2021

A series with same common difference is known as arithmetic series. The first term of series is a and common difference is d. The series is looks like a, a + d, a + 2d, a + 3d, . . . Task is to find the sum of 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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution to find sum of arithmetic series.

C++

// CPP Program to find the sum of arithmetic
// series.
#include<bits/stdc++.h>
using namespace std;

// Function to find sum of series.
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;
}

// Driver function
int main()
{
int n = 20;
float a = 2.5, d = 1.5;
cout<<sumOfAP(a, d, n);
return 0;
}

Java

// JAVA Program to find the sum of
// arithmetic series.

class GFG{

// Function to find sum of series.
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;
}

// Driver function
public static void main(String args[])
{
int n = 20;
float a = 2.5f, d = 1.5f;
System.out.println(sumOfAP(a, d, n));
}
}

/*This code is contributed by Nikita Tiwari.*/

Python

# Python Program to find the sum of
# arithmetic series.

# Function to find sum of series.
def sumOfAP( a, d,n) :
sum = 0
i = 0
while i < n :
sum = sum + a
a = a + d
i = i + 1
return sum

# Driver function
n = 20
a = 2.5
d = 1.5
print (sumOfAP(a, d, n))

# This code is contributed by Nikita Tiwari.

C#

// C# Program to find the sum of
// arithmetic series.
using System;

class GFG {

// Function to find sum of series.
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;
}

// Driver function
public static void Main()
{
int n = 20;
float a = 2.5f, d = 1.5f;

Console.Write(sumOfAP(a, d, n));
}
}

// This code is contributed by parashar.

PHP

<?php
// PHP Program to find the sum
// of arithmetic series.

// Function to find sum of series.
function sumOfAP(\$a, \$d, \$n)
{
\$sum = 0;
for (\$i = 0; \$i < \$n; \$i++)
{
\$sum = \$sum + \$a;
\$a = \$a + \$d;
}
return \$sum;
}

// Driver Code
\$n = 20;
\$a = 2.5; \$d = 1.5;
echo(sumOfAP(\$a, \$d, \$n));

// This code is contributed by Ajit.
?>

Javascript

<script>

// Javascript Program to find the sum of arithmetic
// series.

// Function to find sum of series.
function sumOfAP(a, d, n)
{
let sum = 0;
for (let i=0;i<n;i++)
{
sum = sum + a;
a = a + d;
}
return sum;
}

// Driver function

let n = 20;
let a = 2.5, d = 1.5;
document.write(sumOfAP(a, d, n));

// This code is contributed by Mayank Tyagi

</script>

Output:

335

Time Complexity: O(n)
An Efficient solution to find the sum of arithmetic series is to use below formula.

Sum of arithmetic series
= ((n / 2) * (2 * a + (n - 1) * d))
Where
a - First term
d - Common difference
n - No of terms

C++

// Efficient solution to find sum of arithmetic series.
#include<bits/stdc++.h>
using namespace std;

float sumOfAP(float a, float d, float n)
{
float sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}

// Driver code
int main()
{
float n = 20;
float a = 2.5, d = 1.5;
cout<<sumOfAP(a, d, n);
return 0;
}

Java

// Java Efficient solution to find
// sum of arithmetic series.
class GFG
{
static float sumOfAP(float a, float d, float n)
{
float sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}

// Driver code
public static void main (String[] args)
{
float n = 20;
float a = 2.5f, d = 1.5f;
System.out.print(sumOfAP(a, d, n));
}
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 Efficient
# solution to find sum
# of arithmetic series.

def  sumOfAP(a,  d,  n):
sum = (n / 2) * (2 * a + (n - 1) * d)
return sum

# Driver code
n = 20
a = 2.5
d = 1.5

print(sumOfAP(a, d, n))

# This code is
# contributed by sunnysingh

C#

// C# efficient solution to find
// sum of arithmetic series.
using System;

class GFG {

static float sumOfAP(float a,
float d,
float n)
{
float sum = (n / 2) *
(2 * a +
(n - 1) * d);
return sum;
}

// Driver code
static public void Main ()
{
float n = 20;
float a = 2.5f, d = 1.5f;
Console.WriteLine(sumOfAP(a, d, n));
}
}

// This code is contributed by Ajit.

PHP

<?php
// Efficient PHP code to find sum
// of arithmetic series.

// Function to find sum of series.
function sumOfAP(\$a, \$d, \$n)
{
\$sum = (\$n / 2) * (2 * \$a +
(\$n - 1) * \$d);
return \$sum;
}

// Driver code
\$n = 20;
\$a = 2.5; \$d = 1.5;
echo(sumOfAP(\$a, \$d, \$n));

// This code is contributed by Ajit.
?>

Javascript

// Efficient solution to find sum of arithmetic series.

function sumOfAP(a, d, n) {
let sum = (n / 2) * (2 * a + (n - 1) * d);
return sum;
}

// Driver code
let n = 20;
let a = 2.5, d = 1.5;
document.write(sumOfAP(a, d, n));

// This code is contributed by Ashok
Output
335

Time Complexity: O(1)
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 geometric 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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