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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

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.

∴100 = 5 + (n –1) 5

⇒ 5n = 100

⇒ n = 20

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.

∴100 = 10 + (n –1) (10)

⇒ 100 = 10n

⇒ n = 10

∴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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the sum of natural numbers =n(n+1)/2

there fore the sum of all integer from 1 -100 = (100 X 101)/2 = 5050

from the below pattern we can find out sum of odd number from 1-100

1=12

1+3 = 22

1+3+5=32

the number of odd number from 1to 100 is 50 there for there addition would be 50 square which is = 2500

meritnation did not take much word thus i have writin it in  another post

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the sum of iteger - the sum of odd number =number of even numbers

5050 - 2500= 2550

the sum of even numbers is 2550

the are 10 number  between 1-100 which are not divisibel by 2

this numbers are

5,15,25,35,45,55,,65,75,85,95

meritnation did not take much word thus i have writin it in  another post
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5+15+25+35+45+55+65+75+85+95=

5(1+3+5+7+9+11+13+15+17+19)=

5(10 square)=500

thus the addition  of number divisibel from 2 and 5 from 1 to 100 is 2550+500=3050

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 3050

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