Question

The sum of all natural numbers from 100to300 which are exactly divided by4and 5​

Answer 1

Answer:

Step-by-step explanation:

let's first add natural numbers divisible by 4

first we need to find the numbers between them

so a=100 ,tn=300, d=4

tn=a+(n-1)d

300=100+4n-4

204=4n

n=51

now let's add them

Sn=n/2[2a+(n-1)d]

S51=51/2[2(100)+(51-1)4]

=51/2(400)

=10200

so addition of natural numbers divisible by 4 are 10200

similarly for 5

a=100 tn=300 d=5

tn=a+(n-1)d

300=100+(n-1)5

5n=205

n=41

so let's add the numbers

Sn=n/2[2a+(n-1)d]

S41=41/2[2(100)+(51-1)5]

=41/2(450)

=9225

so the total is 10200+9225

=19425