Average of first n odd naturals numbers

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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 from 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
    sum = 0
    for i in range(0, n) :
        sum = sum + (2 * i + 1)
      
    # Average of first 
    # n odd numbers
    return sum//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.
?>

Javascript

<script>
// javascript 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
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += (2 * i + 1);
 
    // Average of first
    // n odd numbers
    return sum / n;
}
 
// Driver Code
    let n = 20;
    document.write(avg_of_odd_num(n));
 
// This code is contributed by todaysgaurav 
</script>

Output :

Time Complexity: O(n)

Auxiliary Space: O(1)
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.
?>

Javascript

<script>
// javascript 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
    var n = 8;
    document.write(avg_of_odd_num(n));
 
// This code is contributed by gauravrajput1 
</script>

Output :

Time Complexity : O(1)

Space Complexity: O(1) since using constant variables

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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