Find the sum of all natural numbers between 100 and 1000 which are multiples of 5.

Answer

To find the sum of all natural numbers between 100 and 1000 that are multiples of 5, we can follow these steps:

### Step 1: Identify the first and last terms
The first multiple of 5 greater than 100 is 105, and the last multiple of 5 less than 1000 is 995.

### Step 2: Write the sequence
The sequence of multiples of 5 between 100 and 1000 can be written as:
\[ 105, 110, 115, \ldots, 995 \]

### Step 3: Identify the first term (a) and common difference (d)
- First term \( a = 105 \)
- Common difference \( d = 5 \)

### Step 4: Find the number of terms (n)
To find the number of terms in the sequence, we can use the formula for the nth term of an arithmetic sequence:
\[ a_n = a + (n-1)d \]
Setting \( a_n = 995 \):
\[ 995 = 105 + (n-1) \cdot 5 \]
Now, solve for \( n \):
1. Subtract 105 from both sides:
\[ 995 - 105 = (n-1) \cdot 5 \]
\[ 890 = (n-1) \cdot 5 \]
2. Divide both sides by 5:
\[ n - 1 = \frac{890}{5} \]
\[ n - 1 = 178 \]
3. Add 1 to both sides:
\[ n = 179 \]

### Step 5: Use the sum formula for an arithmetic series
The sum \( S_n \) of the first \( n \) terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} (a + l) \]
Where:
- \( n = 179 \)
- \( a = 105 \) (first term)
- \( l = 995 \) (last term)

Substituting the values:
\[ S_{179} = \frac{179}{2} (105 + 995) \]
\[ S_{179} = \frac{179}{2} \cdot 1100 \]
\[ S_{179} = \frac{179 \cdot 1100}{2} \]
\[ S_{179} = 179 \cdot 550 \]
\[ S_{179} = 98450 \]

### Final Answer
The sum of all natural numbers between 100 and 1000 that are multiples of 5 is **98450**.