find the sum of natural numbers between 1 and 100 which are divisible by 3
Answers
Answered by
0
first term=3
last term=99
we write 3,6,9,12.........99
so. tn= a+(n-1)d
99=3+(n-1)3
99=3+3n-3
99=3n
99%3=n
33=n
so tn no.of between 1 to 100 divisble by 3 is 33
last term=99
we write 3,6,9,12.........99
so. tn= a+(n-1)d
99=3+(n-1)3
99=3+3n-3
99=3n
99%3=n
33=n
so tn no.of between 1 to 100 divisble by 3 is 33
Similar questions