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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

 

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))

This article is contributed by Dharmendra kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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