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

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Numbers divisible by 2 from 1 to 100 are $2,4,6................100$

This sequence is an A.P.

Here , first term =a =2

common difference = 2.

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

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

$\Rightarrow \, \, 98 = (n-1)2$

$\Rightarrow \, \, 49 = (n-1)$

$\Rightarrow \, \, n=49+1=50$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{50}{2}[2(2)+(50-1)2]$

$= \frac{50}{2}[2(2)+49(2)]$

$= 25\times 102$

$= 2550$

Numbers divisible by 5 from 1 to 100 are $5,10,15................100$

This sequence is an A.P.

Here , first term =a =5

common difference = 5.

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

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

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

$\Rightarrow \, \, 19 = (n-1)$

$\Rightarrow \, \, n=19+1=20$

$S_n = \frac{n}{2}[2a+(n-1)d]$

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

$= \frac{20}{2}[2(5)+19(5)]$

$= 10\times 105=1050$

Numbers divisible by both 2 and 5 from 1 to 100 are $10,20,30................100$

This sequence is an A.P.

Here , first term =a =10

common difference = 10

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

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

$\Rightarrow \, \, 90 = (n-1)10$

$\Rightarrow \, \, 9 = (n-1)$

$\Rightarrow \, \, n=9+1=10$

$S_n = \frac{n}{2}[2a+(n-1)d]$

$= \frac{10}{2}[2(10)+(10-1)10]$

$= \frac{10}{2}[2(10)+9(10)]$

$= 5\times 110=550$

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

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

Posted by

seema garhwal

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