Average of first n even natural numbers?

The average of first n even natural numbers is the sum of the numbers divided by the count of numbers. The first n even natural numbers are 2, 4, 6, 8, ..., 2n.

Syntax

Average = Sum of first n even numbers / n

You can calculate this using two methods −

• Find the sum of n even natural numbers using a loop and divide by n

• Use the mathematical formula for direct calculation

Method 1: Using Loop

This method iterates through the first n even numbers, calculates their sum, and then finds the average −

#include <stdio.h>

int main() {
    int n = 5;
    int sum = 0;
    float average = 0;
    
    printf("First %d even natural numbers: ", n);
    for (int i = 1; i <= n; i++) {
        int evenNum = i * 2;
        printf("%d ", evenNum);
        sum += evenNum;
    }
    
    average = (float)sum / n;
    printf("\nSum = %d<br>", sum);
    printf("Average = %.2f<br>", average);
    
    return 0;
}

First 5 even natural numbers: 2 4 6 8 10 
Sum = 30
Average = 6.00

Method 2: Using Formula

The mathematical formula for the average of first n even natural numbers is (n + 1). This is derived from the sum formula: Sum = n(n+1) and Average = Sum/n = n+1 −

#include <stdio.h>

int main() {
    int n = 5;
    int average = n + 1;
    
    printf("Using formula: Average of first %d even natural numbers<br>", n);
    printf("Formula: Average = n + 1 = %d + 1 = %d<br>", n, average);
    printf("Average = %d<br>", average);
    
    return 0;
}

Using formula: Average of first 5 even natural numbers
Formula: Average = n + 1 = 5 + 1 = 6
Average = 6

Comparison

Conclusion

The formula method (n+1) is more efficient for calculating the average of first n even natural numbers, especially for larger values of n. The loop method helps understand the underlying concept but becomes slower as n increases.