Question

Find the sum of all natural numbers between 250 nd 1000 which are exactly divisible by 3

Answer 1

In AP

A
a1 =252
an=999
n=?
an=a+(n-1)d
999=252+(n-1)3
999-252=3(n-1)
747/3=n-1
249+1=n
n=250
by using Sn=[(n/2)2a+(n-1)d]
= [(250/2)2×252+(250-1)3]
= 125 ×504+(249×3)
=156375


hope it's help u!