# Average of first n odd naturals numbers

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

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 ` `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

 ` `

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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