Find the sum of the multiples of 3,between 1and 100
Answers
Answered by
4
a=3
d=3
l=99
l=a+(n-1)d
99=3+(n-1)3
96=(n-1)3
32=n-1
n=33
Sum=n/2(a+l)
=33/2(99+3)
=1683
d=3
l=99
l=a+(n-1)d
99=3+(n-1)3
96=(n-1)3
32=n-1
n=33
Sum=n/2(a+l)
=33/2(99+3)
=1683
Similar questions