Question

The sum of all the natural numbers from 100 and 300,which is divisible by 5 is​

Answer 1

Answer:

set of numbers which are divisible by 5 are

100,105,110......300

Now

last term = a+(n-1)d

300=100+(n-1)5

200/5=n-1

20+1=n

n=21

Now

sum = n/2(a+an)

sum = 21/2(100+300)

sum = 21/2(400)

sum = 21×200

sum = 4200

Hence sum of all natural numbers lying from 100 to 300 which are divisible by 5 is 4200