Approach

1. Used a for loop starting from i=a.

2. Increased i by d each time.

3. Stoped when value of i exceeds b.

4. If the number is even (i % 2 == 0), add it to evenSum.

5. Print the final sum.

6. Time Complexity -> O((p) where p=(b-a)/d

7. Space Complexity -> O(1)