Question

C. G Find the sum of all multiples of 3 natural numbers between 1 and100

Answer 1

Answer:

the multiples of 3 between 1 and 100 are

3, 6, .....99

therefore, a=3

an=99

d=3

first we'll find out no. of terms

An=a+(n-1)d

99=3+(n-1)3

99-3=3n-3

96-3+3=3n

n=96/3 = 32

so we got no. of terms now we can findout sum of 32 terms

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

=32/2(2*3+(32-1)3)

=16(6+93)

=16*99

=1584

so the sum of all multiples of 3 between 1 and 100 is 1584

hope it helps