Solution:

To find the sum of integers from 1 to 100 which are divisible by 2 or 5, we need to find the sum of integers which are divisible by 2 and the sum of integers which are divisible by 5 and then subtract the sum of integers which are divisible by both 2 and 5 because they are counted twice.

Sum of integers divisible by 2:

The first integer divisible by 2 is 2 and the last integer divisible by 2 is 100. So, we need to find the sum of an arithmetic progression with a common difference of 2, whose first term is 2 and last term is 100.

Let the number of terms in this progression be n. So, we need to find n such that 100 = 2 + (n-1)2. Solving for n, we get n = 50.

So, the sum of integers divisible by 2 is given by

S1 = 2 + 4 + 6 + ... + 100

= 2(1 + 2 + 3 + ... + 50)

= 2(50)(51)/2

= 2550

Sum of integers divisible by 5:

The first integer divisible by 5 is 5 and the last integer divisible by 5 is 100. So, we need to find the sum of an arithmetic progression with a common difference of 5, whose first term is 5 and last term is 100.

Let the number of terms in this progression be m. So, we need to find m such that 100 = 5 + (m-1)5. Solving for m, we get m = 20.

So, the sum of integers divisible by 5 is given by

S2 = 5 + 10 + 15 + ... + 100

= 5(1 + 2 + 3 + ... + 20)

= 5(20)(21)/2

= 1050

Sum of integers divisible by both 2 and 5:

The first integer divisible by both 2 and 5 is 10 and the last integer divisible by both 2 and 5 is 100. So, we need to find the sum of an arithmetic progression with a common difference of 10, whose first term is 10 and last term is 100.

Let the number of terms in this progression be k. So, we need to find k such that 100 = 10 + (k-1)10. Solving for k, we get k = 10.

So, the sum of integers divisible by both 2 and 5 is given by

S3 = 10 + 20 + ... + 100

= 10(1 + 2 + ... + 10)

= 10(10)(11)/2

= 550

Therefore, the sum of integers from 1 to 100 which are divisible by 2 or 5 is

S = S1 + S2 - S3

= 2550 + 1050 - 550

= 3050

Hence, the correct answer is option D.