Average of first n even natural numbers
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++
// 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 numbers 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 int main() { int n = 9; cout << avg_of_even_num(n); return 0; } |
Java
// java implementation to find Average // of sum of first n natural even number import 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
# Python3 implementation to # find Average of sum of # first n natural even # number # Function to find average # of sum of first n even # numbers def 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 = 9 print (avg_of_even_num(n)) # This code is contributed by upendra singh bartwal |
C#
// C# implementation to find // Average of sum of first // n natural even number using 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 // PHP implementation to find Average // of sum of first n natural even numbers // function to find average of // sum of first n even numbers function 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. ?> |
Javascript
<script> // javascript implementation to find Average // of sum of first n natural even numbers // function to find average of // sum of first n even numbers function 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)
Auxiliary Space: O(1) as it is using constant space
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++
// 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 numbers int avg_of_even_num( int n) { return n+1; } // Driver Code int main() { int n = 8; cout << avg_of_even_num(n) << endl; return 0; } |
Java
// Java Program to find the average // of sum of first n even numbers import 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 |
Python3
# Python 3 Program to # find the average # of sum of first n # even numbers # Return the average of sum # of first n even numbers def avg_of_even_num(n) : return n + 1 # Driven Program n = 8 print (avg_of_even_num(n)) # This code is contributed # by Nikita Tiwari. |
C#
// C# Program to find the average // of sum of first n even numbers using 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 // PHP Program to find the average // of sum of first n even numbers // Return the average of sum // of first n even numbers function avg_of_even_num( $n ) { return $n + 1; } // Driver Code $n = 8; echo (avg_of_even_num( $n )); // This code is contributed by Ajit. ?> |
Javascript
<script> // javascript Program to find the average // of sum of first n even numbers // Return the average of sum // of first n even numbers function 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)
Auxiliary Space: O(1)
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