The sum of all natural numbers between 100 and 1000 which are multiples of 5

To find the sum of all natural numbers between 100 and 1000 that are multiples of 5, we can use the concept of arithmetic progression and sum of an arithmetic series.

Arithmetic Progression:
An arithmetic progression is a sequence of numbers in which the difference between any two consecutive terms is constant. In this case, the common difference is 5 because we are considering multiples of 5.

Sum of an Arithmetic Series:
The sum of an arithmetic series can be calculated using the formula:
S = (n/2)(2a + (n-1)d),
where S is the sum of the series, n is the number of terms, a is the first term, and d is the common difference.

Step 1: Find the first term (a):
The first term of the arithmetic progression will be the smallest multiple of 5 within the given range. In this case, it is 100 since it is the closest multiple of 5 greater than or equal to 100.

Step 2: Find the last term (l):
The last term of the arithmetic progression will be the largest multiple of 5 within the given range. In this case, it is 1000 since it is the closest multiple of 5 less than or equal to 1000.

Step 3: Find the number of terms (n):
The number of terms can be calculated using the formula:
n = (l - a)/d + 1,
where l is the last term, a is the first term, and d is the common difference.

In this case, the number of terms is (1000 - 100)/5 + 1 = 181.

Step 4: Calculate the sum (S):
Using the formula for the sum of an arithmetic series mentioned earlier, we can calculate the sum of all the natural numbers between 100 and 1000 that are multiples of 5.

S = (181/2)(2 * 100 + (181-1) * 5)
= 90 * (200 + 900)
= 90 * 1100
= 99,000

Therefore, the sum of all natural numbers between 100 and 1000 that are multiples of 5 is 99,000.