Find the sum of integers from $1$ to $100$ that are divisible by $2$ or $5.$
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 $2,4,6 \ldots \ldots 100$
This forms an $A.P.$ with both the first term and common difference equal to $2.$
$\Rightarrow 100=2+(n-1) 2$
$\Rightarrow n=50$
$\therefore 2+4+6+\ldots \ldots+100=\frac{50}{2}[2(2)+(50-1)(2)]$
$=\frac{50}{2}[4+98]$
$=(25)(102)$
$=2550$
The integers from $1$ to $100 ,$ which are divisible by $5,10 \ldots . .100$
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 $10,20, \ldots \ldots 100$
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 $3050$
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