Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

First let’s find the integers from 1 to 100, which are divisible by 2.

And, they are 2, 4, 6… 100.

Clearly, this forms an A.P. with the first term and common difference both equal to 2.

So, we have

100 = 2 + (n –1) 2

n = 50

Hence, the sum is

\( 2+4+6+...+100\)

= \(\dfrac{50}{2}[2(2)+(50-1)(2)]\)

= \(\dfrac{50}{2}[4+8]\)

= \( (25)(102)\)

= 2550

Now, the integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This also forms an A.P. with the first term and common difference both equal to 5.

So, we have

100 = 5 + (n –1) 5

5n = 100

n = 20

Hence, the sum is

5+10+....+100

= \( \dfrac{20}{2}[2(5)+(20-1)5]\)

= \( 10[10+(19)5]\)

= \( 10[10+195]\)

\(= 10\times 105=1050\)

Lastly, the integers which are divisible by both 2 and 5, are 10, 20, … 100.

And this also forms an A.P. with the first term and common difference both equal to 10.

So, we have

100 = 10 + (n –1) (10)

100 = 10n

n = 10

10+20+........+100

\( \dfrac{10}{2}[2(10)+(10-1)(10)]\)

= 5[20 + 90]

= 5(110)=550

Thus, the required sum = 2550 + 1050 – 550 = 3050

Therefore, the sum of the integers from 1 to 100, which are divisible by 2 or 5, is 3050.