Question

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

Answer 1

Answer:

2,4,6,8......100

Step-by-step explanation:

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Answer 2

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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.

$\tt ⇒ 100=2+(n-1)2$

$\tt ⇒ n= 50$

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

$\tt 2+4+6+…+100 = (50/2)[2(2)+(50-1)(2)]$

$\tt = (50/2)(4+98)$

$\tt = 25(102)$

$\tt = 2550$

The integers from 1 to 100, which are divisible by 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

$\tt ⇒5n = 100$

$\tt ⇒ n= 100/5$

$\tt ⇒ n= 20$

Therefore, the sum of integers from 1 to 100 that are divisible by 2 is given as:

$\tt 5+10+15+…+100= (20/2)[2(5)+(20-1)(5)]$

$\tt = (20/2)(10+95)$

$\tt = 10(105)$

$\tt = 1050$

Hence, the integers from 1 to 100, 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

$\tt ⇒10n = 100$

$\tt ⇒ n= 100/10$

$\tt ⇒ n= 10$

$\tt 10+20+…+100= (10/2)[2(10)+(10-1)(10)]$

$\tt = (10/2)(20+90)$

$\tt = 5(110)$

$\tt = 550$

Therefore, the required sum is:

$\tt = 2550+ 1050 – 550$

$\tt = 3050$

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

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