## Answers

**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**