Question

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

Answer 1

The integers from 1 to 100 which are divisible by 2 are:

2, 4, 6, .... 100

Here a = 2, d = 4 - 2 = 2 and $a_n=100$

$\because a_n=a+(n-1)d$

$\therefore 100=2+(n-1)2$

$\Rightarrow 100-2=(n-1{)}^2$

$\Rightarrow (n-1)=\frac{98}{2}=49$

$\therefore n=49+1=50$

Now, $S_1=\frac{50}{2}[2+100]$

$=25\times 102=2550$

Now, the integers from 1 to 100 which are divisible by 5 are 5, 10, 15, ...., 100.

Here, a = 5, d = 10 - 5 = 5 and $a_n=100$

$\because a_n=a+(n-1)d$

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

$\Rightarrow 100-5=(n-1)\times 5$

$\Rightarrow 95=(n-1)\times 5$

$\Rightarrow (n-1)=\frac{95}{5}=19\Rightarrow n=20$

$\therefore S_2=\frac{20}{2}[5+100]$

$=10\times 105=1050$

Also, the integers from 1 to 100 which are divisible by both 2 and 5 are :

10, 20, 30, ...., 100

Here, a = 10, d = 20 - 10 = 10 and $a_n=100$

Since, $a_n=a+(n-1)d$

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

$\Rightarrow n=9+1=10$

Now, $S_{10}=\frac{10}{2}[10+100]$

$=5\times 110=550$

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

$(S_1+S_2)-S_3$

i.e. $(2550+1050)-550=3050$.