Average of first n even natural numbers - GeeksforGeeks

24 Mar, 2021
Given a number n then Find the Average of first n even natural numbers Ex.= 2 + 4 + 6 + 8 + 10 + 12 +.........+ 2n. Examples :  
Input  : 7
Output : 8
(2 + 4 + 6 + 8 + 10 + 12 + 14)/7 = 8 

Input  : 5
Output : 6
(2 + 4 + 6 + 8 + 10)/5 = 6
 
Naive Approach:- In this program iterate the loop , finding total sum of first n even numbers and divided by n.it take 0(N) time. 
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation to find Average// of sum of first n natural even numbers#include <bits/stdc++.h>using namespace std; // function to find average of// sum of first n even numbersint avg_of_even_num(int n){    // sum of first n even numbers    int sum = 0;    for (int i = 1; i <= n; i++)        sum += 2*i;     // calculating Average    return sum/n;} // Driver Codeint main(){    int n = 9;    cout << avg_of_even_num(n);    return 0;}
// java implementation to find Average// of sum of first n natural even numberimport java.io.*; class GFG {         // function to find average of    // sum of first n even numbers    static int avg_of_even_num(int n)    {         // sum of first n even numbers    int sum = 0;              for (int i = 1; i <= n; i++)        sum += 2*i;     // calculating Average    return (sum / n);    }    public static void main (String[] args) {         int n = 9;    System.out.print(avg_of_even_num(n));                 }} // this code is contributed by 'vt_m'
# Python3 implementation to# find Average of sum of# first n natural even# number # Function to find average# of sum of first n even# numbersdef avg_of_even_num(n):         # sum of first n even    # numbers    sum=0    for i in range(1, n + 1):        sum=sum + 2 * i         # calculating Average    return sum / n n=9print(avg_of_even_num(n)) # This code is contributed by upendra singh bartwal
// C# implementation to find// Average of sum of first// n natural even numberusing System; class GFG {         // function to find average of    // sum of first n even numbers    static int avg_of_even_num(int n)    {         // sum of first n even numbers    int sum = 0;         for (int i = 1; i <= n; i++)        sum += 2 * i;     // calculating Average    return (sum / n);    }         // driver code    public static void Main () {         int n = 9;    Console.Write(avg_of_even_num(n));                 }} // This code is contributed by 'vt_m'
<?php// PHP implementation to find Average// of sum of first n natural even numbers // function to find average of// sum of first n even numbersfunction avg_of_even_num($n){    // sum of first n even numbers    $sum = 0;    for ($i = 1; $i <= $n; $i++)        $sum += 2 * $i;     // calculating Average    return $sum / $n;} // Driver Code$n = 9;echo(avg_of_even_num($n)); // This code is contributed by Ajit.?>
<script>// javascript implementation to find Average// of sum of first n natural even numbers // function to find average of// sum of first n even numbersfunction avg_of_even_num( n){     // sum of first n even numbers    let sum = 0;    for (let i = 1; i <= n; i++)        sum += 2*i;     // calculating Average    return sum/n;} // Driver Code    let n = 9;    document.write(avg_of_even_num(n)); // This code is contributed by todaysgaurav </script>
Output :  
  10
Time Complexity : O(N) Method 2 :- The idea is the sum of first n even number is n(n+1), for find the Average of first n even numbers divide by n, hence formula is n(n + 1) / n = ( n + 1). i.e. Average of first n even numbers is n+1. it take 0(1) time.  
                  Avg of sum of N even natural number = (N + 1)
Proof  
Sum of first n terms of an A.P.(Arithmetic Progression)
= (n/2) * [2*a + (n-1)*d].....(i)
where, a is the first term of the series and d is
the difference between the adjacent terms of the series.

Here, a = 2, d = 2, applying these values to eq.(i), get
Sum = (n/2) * [2*2 + (n-1)*2]
    = (n/2) * [4 + 2*n - 2]
    = (n/2) * (2*n + 2)
    = n * (n + 1)

 finding the Avg so divided by n = n*(n+1)/n
                                      = (n+1)
 
C++
Java
Python3
C#
PHP
Javascript
// CPP Program to find the average// of sum of first n even numbers#include <bits/stdc++.h>using namespace std; // Return the average of sum// of first n even numbersint avg_of_even_num(int n){    return n+1;}     // Driver Codeint main(){    int n = 8;    cout << avg_of_even_num(n) << endl;    return 0;}
// Java Program to find the average// of sum of first n even numbersimport java.io.*; class GFG{     // Return the average of sum    // of first n even numbers    static int avg_of_even_num(int n)    {        return n + 1;    }         public static void main (String[] args) {                 int n = 8;        System.out.println(avg_of_even_num(n));             }} // This code is contributed by vt_m
# Python 3 Program to# find the average# of sum of first n# even numbers # Return the average of sum# of first n even numbersdef avg_of_even_num(n) :         return n+1           # Driven Programn = 8print(avg_of_even_num(n))  # This code is contributed# by Nikita Tiwari.
// C# Program to find the average// of sum of first n even numbersusing System; class GFG {     // Return the average of sum    // of first n even numbers    static int avg_of_even_num(int n)    {        return n + 1;    }         // driver code       public static void Main () {                 int n = 8;        Console.Write(avg_of_even_num(n));             }} // This code is contributed by vt_m
<?php// PHP Program to find the average// of sum of first n even numbers // Return the average of sum// of first n even numbersfunction avg_of_even_num($n){    return $n + 1;}     // Driver Code$n = 8;echo(avg_of_even_num($n)); // This code is contributed by Ajit.?>
<script>// javascript Program to find the average// of sum of first n even numbers // Return the average of sum// of first n even numbersfunction avg_of_even_num(n){    return n + 1;}var n = 8;document.write(avg_of_even_num(n)); // This code is contributed by Amit Katiyar</script>
Output:  
9
Time Complexity : O(1)  
jit_t
todaysgaurav
amit143katiyar
series
Mathematical
Mathematical
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