Find the sum of first 100 even natural numbers which are divisible by 5.

Answer

To find the sum of the first 100 even natural numbers that are divisible by 5, we can follow these steps:

### Step 1: Identify the first 100 even natural numbers
The first 100 even natural numbers are:
\[ 2, 4, 6, 8, \ldots, 200 \]
These numbers can be expressed as \( 2n \) where \( n \) ranges from 1 to 100.

### Step 2: Identify even natural numbers that are divisible by 5
The even natural numbers that are divisible by 5 can be found by identifying the multiples of 10 (since the least common multiple of 2 and 5 is 10). The sequence of even natural numbers divisible by 5 starts from 10 and continues as follows:
\[ 10, 20, 30, 40, \ldots \]
This can be expressed as:
\[ 10n \]
where \( n \) is a natural number.

### Step 3: Determine the first 100 terms of this sequence
The first 100 terms of the sequence \( 10, 20, 30, \ldots \) can be written as:
\[ 10 \times 1, 10 \times 2, 10 \times 3, \ldots, 10 \times 100 \]
This gives us:
\[ 10, 20, 30, 40, \ldots, 1000 \]

### Step 4: Identify the arithmetic progression (AP)
The sequence \( 10, 20, 30, \ldots, 1000 \) is an arithmetic progression (AP) where:
- The first term \( a = 10 \)
- The common difference \( d = 10 \)
- The number of terms \( n = 100 \)

### Step 5: Use the formula for the sum of an AP
The sum \( S_n \) of the first \( n \) terms of an arithmetic progression can be calculated using the formula:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
Substituting the known values:
- \( n = 100 \)
- \( a = 10 \)
- \( d = 10 \)

### Step 6: Calculate the sum
Substituting these values into the formula:
\[ S_{100} = \frac{100}{2} \times (2 \times 10 + (100 - 1) \times 10) \]
\[ = 50 \times (20 + 99 \times 10) \]
\[ = 50 \times (20 + 990) \]
\[ = 50 \times 1010 \]
\[ = 50500 \]

### Conclusion
The sum of the first 100 even natural numbers that are divisible by 5 is:
\[ \boxed{50500} \]