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

Explanation

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 sequence of even natural numbers divisible by 5.
The first few even natural numbers that are divisible by 5 are:
- 10 (5 × 2)
- 20 (5 × 4)
- 30 (5 × 6)
- 40 (5 × 8)
- ...

We can see that these numbers form an arithmetic progression (AP) where the first term (A) is 10 and the common difference (d) is also 10.

### Step 2: Determine the last term of the sequence.
Since we need the first 100 terms, we can find the last term (L) of this sequence:
- The nth term of an AP can be found using the formula:
\[ L = A + (n - 1) \cdot d \]
- Here, \( n = 100 \), \( A = 10 \), and \( d = 10 \):
\[ L = 10 + (100 - 1) \cdot 10 = 10 + 99 \cdot 10 = 10 + 990 = 1000 \]

### Step 3: Use the formula for the sum of the first n terms of an AP.
The sum \( S_n \) of the first n terms of an AP can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (A + L) \]
- Substituting the values \( n = 100 \), \( A = 10 \), and \( L = 1000 \):
\[ S_{100} = \frac{100}{2} \cdot (10 + 1000) = 50 \cdot 1010 = 50500 \]

### Conclusion:
The sum of the first 100 even natural numbers that are divisible by 5 is **50500**.

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