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Average of first n odd naturals numbers
  • Last Updated : 24 Apr, 2018

Given a Number n then find the Average of first n odd numbers
1 + 3 + 5 + 7 + 9 +………….+ (2n – 1)

Examples :

Input  : 5
Output : 5
(1 + 3 + 5 + 7 + 9)/5 = 5 

Input  : 10
Output : 10
(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)/10 =10

Method 1 ( Naive Approach:)
A simple solution is to iterate loop form 1 to n time. Through sum of all odd numbers and divided by n.This solution take O(N) time.

C++




// A  C++ program to find average of
// sum of first n odd natural numbers.
#include <iostream>
using namespace std;
  
// Returns the Avg of
// first n odd numbers
int avg_of_odd_num(int n)
{
  
    // sum of first n odd number
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += (2 * i + 1);
  
    // Average of first
    // n odd numbers
    return sum / n;
}
  
// Driver Code
int main()
{
    int n = 20;
    cout << avg_of_odd_num(n);
    return 0;
}


Java




// Java program to find average of
// sum of first n odd natural numbers.
import java.io.*;
  
class GFG {
  
    // Returns the Avg of
    // first n odd numbers
    static int avg_of_odd_num(int n)
    {
  
        // sum of first n odd number
        int sum = 0;
  
        for (int i = 0; i < n; i++)
            sum += (2 * i + 1);
  
        // Average of first
        // n odd numbers
        return sum / n;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        int n = 20;
        avg_of_odd_num(n);
  
        System.out.println(avg_of_odd_num(n));
    }
}
  
// This code is contributed by vt_m


Python3




# A Python 3 program
# to find average of
# sum of first n odd
# natural numbers.
  
# Returns the Avg of
# first n odd numbers
def avg_of_odd_num(n) :
  
    # sum of first n odd number
    sm = 0
    for i in range(0, n) :
        sm = sm + (2 * i + 1)
       
    # Average of first 
    # n odd numbers
    return sm//n
  
   
# Driver Code
n = 20
print(avg_of_odd_num(n))
  
  
# This code is contributed
# by Nikita Tiwari.


C#




// C# program to find average
// of sum of first n odd
// natural numbers.
using System;
  
class GFG {
  
    // Returns the Avg of
    // first n odd numbers
    static int avg_of_odd_num(int n)
    {
  
        // sum of first n odd number
        int sum = 0;
  
        for (int i = 0; i < n; i++)
            sum += (2 * i + 1);
  
        // Average of first
        // n odd numbers
        return sum / n;
    }
  
    // Driver code
    public static void Main()
    {
  
        int n = 20;
        avg_of_odd_num(n);
  
        Console.Write(avg_of_odd_num(n));
    }
}
  
// This code is contributed by
// Smitha Dinesh Semwal


PHP




<?php
// A PHP program to find average of
// sum of first n odd natural numbers.
  
// Returns the Avg of
// first n odd numbers
function avg_of_odd_num($n)
{
  
    // sum of first n odd number
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        $sum += (2 * $i + 1);
  
    // Average of first
    // n odd numbers
    return $sum / $n;
}
  
// Driver Code
$n = 20;
echo(avg_of_odd_num($n));
  
// This code is contributed by Ajit.
?>


Output :



 20

Time Complexity : O(n)

Method 2 (Efficient Approach:)
The idea is the sum of first n odd number is n2, for find the Average of first n odd numbers so it is divide by n, hence formula is n2/n = n. it take O(1) time.

                           Avg of sum of first N odd Numbers = N

C++




// CPP Program to find the average
// of sum of first n odd numbers
#include <bits/stdc++.h>
using namespace std;
  
// Return the average of sum
// of first n odd numbers
int avg_of_odd_num(int n)
{
    return n;
}
  
// Driver Code
int main()
{
    int n = 8;
    cout << avg_of_odd_num(n);
    return 0;
}


Java




// java Program to find the average
// of sum of first n odd numbers
import java.io.*;
  
class GFG {
  
    // Return the average of sum
    // of first n odd numbers
    static int avg_of_odd_num(int n)
    {
        return n;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        int n = 8;
  
        System.out.println(avg_of_odd_num(n));
    }
}
  
// This code is contributed by vt_m


Python3




# Python 3 Program to
# find the average
# of sum of first n
# odd numbers
  
# Return the average of sum
# of first n odd numbers
def avg_of_odd_num(n) :
    return n
      
  
# Driver Code
n = 8
print(avg_of_odd_num(n))
  
  
# This code is contributed
# by Nikita Tiwari.


C#




// C# Program to find the average
// of sum of first n odd numbers
using System;
  
class GFG {
    // Return the average of sum
    // of first n odd numbers
    static int avg_of_odd_num(int n)
    {
        return n;
    }
  
    // Driver Code
    public static void Main()
    {
        int n = 8;
        Console.Write(avg_of_odd_num(n));
    }
}
// This code is contributed by
// Smitha Dinesh Semwal


PHP




<?php
// PHP Program to find the average
// of sum of first n odd numbers
  
// Return the average of sum
// of first n odd numbers
function avg_of_odd_num($n)
{
    return $n;
}
  
// Driver Code
$n = 8;
echo(avg_of_odd_num($n));
  
// This code is contributed by Ajit.
?>


Output :

 8

Time Complexity : O(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 = 1, d = 2, applying these values to e. q., 
(i), we get
Sum = (n/2) * [2*1 + (n-1)*2]
    = (n/2) * [2 + 2*n - 2]
    = (n/2) * (2*n)
    = n*n
    = n2

Avg of first n odd numbers = n2/n
                           = n

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