Sum of Natural Numbers Divisible by 3 between 250 and 1000
To find the sum of all natural numbers between 250 and 1000 that are exactly divisible by 3, we need to first determine the first and last numbers in this range that are divisible by 3.

Finding the First Number:
The first number that is divisible by 3 and lies between 250 and 1000 is 252. To find this, we divide 250 by 3 and round up to the nearest whole number, which gives us 84. Then, we multiply 84 by 3 to get 252.

Finding the Last Number:
The last number that is divisible by 3 and lies between 250 and 1000 is 999. To find this, we divide 1000 by 3 to get 333.33. Since we are looking for a whole number, we take the floor of 333.33, which gives us 333. Then, we multiply 333 by 3 to get 999.

Calculating the Sum:
Now that we have the first and last numbers, we can calculate the sum of the arithmetic progression. The formula for the sum of an arithmetic progression is given by:
Sum = n/2 * (first term + last term)
In this case, the first term is 252, the last term is 999, and n is the number of terms in the sequence. To find n, we can use the formula for the number of terms in an arithmetic progression:
Number of terms = (last term - first term)/common difference + 1
The common difference in this case is 3, since we are dealing with numbers that are divisible by 3. By substituting the values into the formula, we get the number of terms as 250.
Therefore, the sum of all natural numbers between 250 and 1000 that are exactly divisible by 3 is 125625.