Program for sum of arithmetic series - GeeksforGeeks

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++
Java
Python
C#
PHP
Javascript
// 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 functionint 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 functionn = 20a = 2.5d = 1.5print (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)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++
Java
Python3
C#
PHP
Javascript
// 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 codeint 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 = 20a = 2.5d = 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 codelet 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)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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