Answers

2015-01-04T17:00:35+05:30


2,4,6......,100 are the integers(from 1 to 100) divisible by 2

This forms an A.P series whose first term is 2 and the common difference is equal to 4-2 or 6-4 = 2

we know that nth term of an A.P , Tnth = a +(n-1)d

implies 100=2+(n-1)2

therefore n=50

Sum of n terms of an A.P, Sn = n/2[2a + (n-1)d]

=50/2[2*2 +(50-1)2]

25[4+49*2]

=2550

5,10.....100 are the integers from 1 to 100 that are divisible by 5

This forms an A.P series whose first term is 5 and common difference is 10-5=5

Tnth=a+(n-1)d

100=5+(n-1)5

therefore n=20

Sn= 20/2[2*5 +(20-1)5]

=1050

The integers which are divisible by both 2 and 5 are 10, 20,.....100 This forms an A.P with 10 as first term and common difference is 10

100=10+(n-1)10

Therefore n=10

Sn= 10/2[2*10+(10-1)10]

=550

Required Sum (of the integers from 1 to 100 that are divisible by 2 or 5)=2550+1050-550=3050


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