# Java Program to Find the Sum of First N Odd & Even Numbers

• Last Updated : 19 Oct, 2022

When any number which ends with 0,2,4,6,8 is divided by 2 that is an even number. And when any number ends with 1,3,5,7,9 is not divided by two is an odd number.

Example:

```Input : 8
Output: Sum of First 8 Even numbers = 72
Sum of First 8 Odd numbers = 64```

Approach #1: Iterative

1. Create two variables evenSum and oddSum and initialize them by 0.
2. Start For loop from 1 to 2*n.
3. If i is even Add i with evenSum.
4. Else add i with oddSum.
5. Print evenSum and oddSum at the end of loop.

Below is the implementation of the Java program:

## Java

 `// Calculate the Sum of First N Odd & Even Numbers in Java` `import` `java.io.*;`   `public` `class` `GFG {`   `    ``// Driver function` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``8``;` `        ``int` `evenSum = ``0``;` `        ``int` `oddSum = ``0``;`   `        ``for` `(``int` `i = ``1``; i <= ``2` `* n; i++) {` `            ``// check even & odd using Bitwise AND operator` `            ``if` `((i & ``1``) == ``0``)` `                ``evenSum += i;` `            ``else` `                ``oddSum += i;` `        ``}` `        ``// Sum of even numbers less than 17` `        ``System.out.println(``"Sum of First "` `+ n` `                           ``+ ``" Even numbers = "` `+ evenSum);`   `        ``// sum of odd numbers less than 17` `        ``System.out.println(``"Sum of First "` `+ n` `                           ``+ ``" Odd numbers = "` `+ oddSum);` `    ``}` `}`

Output

```Sum of First 8 Even numbers = 72
Sum of First 8 Odd numbers = 64```

Time Complexity: O(N), where N is the number of First N even/odd numbers.
Auxiliary Space: O(1)

Method 2: Using AP Formulas.

• Sum of First N Even Numbers = n * (n+1)
• Sum of First N Odd Numbers = n * n

Below is the implementation of the above approach:

## Java

 `// Calculate the Sum of First N Odd & Even Numbers in Java` `import` `java.io.*;`   `public` `class` `GFG {`   `    ``// Function to find the sum of even numbers` `    ``static` `int` `sumOfEvenNums(``int` `n) { ``return` `n * (n + ``1``); }`   `    ``// Function to find the sum of odd numbers.` `    ``static` `int` `sumOfOddNums(``int` `n) { ``return` `n * n; }`   `    ``// Driver function` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``10``;` `        ``int` `evenSum = sumOfEvenNums(n);` `        ``int` `oddSum = sumOfOddNums(n);`   `        ``// Sum of even numbers` `        ``System.out.println(``"Sum of First "` `+ n` `                           ``+ ``" Even numbers = "` `+ evenSum);`   `        ``// sum of odd numbers` `        ``System.out.println(``"Sum of First "` `+ n` `                           ``+ ``" Odd numbers = "` `+ oddSum);` `    ``}` `}`

Output

```Sum of First 10 Even numbers = 110
Sum of First 10 Odd numbers = 100```

Time Complexity: O(1)
Auxiliary Space: O(1)

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