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

Explanation

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

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

### Step 2: Identify the common difference
Since we are dealing with multiples of 5, the common difference (d) is 5.

### Step 3: Determine the number of terms (n) in the sequence
We can use the formula for the nth term of an arithmetic progression (AP):
\[ A_n = A + (n - 1)d \]
Where:
- \(A_n\) is the last term (995),
- \(A\) is the first term (105),
- \(d\) is the common difference (5).

Setting up the equation:
\[ 995 = 105 + (n - 1) \times 5 \]
Subtract 105 from both sides:
\[ 890 = (n - 1) \times 5 \]
Now, divide by 5:
\[ n - 1 = \frac{890}{5} = 178 \]
Adding 1 to both sides gives:
\[ n = 179 \]

### Step 4: Use the formula for the sum of an arithmetic series
The sum \(S_n\) of the first n terms of an arithmetic series can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2A + (n - 1)d) \]
Substituting the values we found:
- \(n = 179\)
- \(A = 105\)
- \(d = 5\)

Calculating:
\[ S_{179} = \frac{179}{2} \times (2 \times 105 + (179 - 1) \times 5) \]
Calculating \(2 \times 105\):
\[ 2 \times 105 = 210 \]
Calculating \((179 - 1) \times 5\):
\[ (179 - 1) \times 5 = 178 \times 5 = 890 \]
Now substituting back into the sum formula:
\[ S_{179} = \frac{179}{2} \times (210 + 890) \]
Calculating \(210 + 890\):
\[ 210 + 890 = 1100 \]
Now substituting:
\[ S_{179} = \frac{179}{2} \times 1100 \]
Calculating:
\[ S_{179} = 179 \times 550 = 98450 \]

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

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