Program for sum of arithmetic series

Try it on GfG Practice

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. 

// 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 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 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# 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 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.
?>

<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)

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:

// 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 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 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# 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
// 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.
?>

// 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

335

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))