Question

find the sum of all natural numbers upto 100 which are divisible by 3?​

Answer 1

Hey,

the first natural no. is 3 and the last no. is 99. we ca write those numbers as 3,6,9,12........,99

sum= 3+6+9+......+99

this is an arithmetic progression

here,

first term, a=3

last term, L=99

difference between two consecutive terms, d=3

now lets find the no. of terms 'n'

use formula

L=a+(n-1)d

99=3+(n-1)3

99=3n

or n=33

now lets find the sum:

sum of an arithmetic progression is,

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

=33[2*3+(33-1)3]/2

=33[6+96]/2

=33[102]/2

=33*51

=1683

hope this helps you