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 numbersint 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 Codeint 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 numbersdef 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 Coden = 20print(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 numbersfunction 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 numbersfunction 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 :
20
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 numbersint avg_of_odd_num(int n){ return n;}// Driver Codeint 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 numbersimport 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 numbersdef avg_of_odd_num(n) : return n # Driver Coden = 8print(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 numbersusing 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 numbersfunction 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 :
8
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
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
