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

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

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

⇒100 = 2 + (n –1) 2

⇒ n = 50

$\therefore 2+4+6+\ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]$

$=\frac{50}{2}[4+98]$

$=(25)(102)$

$=2550$

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

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

$\therefore 100=5+(n-1) 5$

$\Rightarrow 5 n=100$

$\Rightarrow n=20$

$\therefore 5+10+\ldots+100=\frac{20}{2}[2(5)+(20-1) 5]$

$=10[10+(19) 5]$

$=10[10+95]=10 \times 105$

$=1050$

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

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

$\therefore 100=10+(n-1)(10)$

$\Rightarrow 100=10 n$

$\Rightarrow n=10$

$\therefore 10+20+\ldots+100=\frac{10}{2}[2(10)+(10-1)(10)]$

$=5[20+90]=5(110)=550$

$\therefore$ Required sum $=2550+1050-550=3050$

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