Question

Find the sum of all natural numbers between 100 and 200

Answer 1

let the numbers are 101,102,103,.........,199
d=t2 - t1 = t3 - t2 = 102-101= 1
these are in AP
a=101
tn = 199
d=1
tn = a+(n-1)d=199
       101+(n-1)1=199
       101+n-1=199
       n=99
Sn= n/2(a+l)
      = 99/2(101+199)
       = 99/2(300)
       = 14850
            

Answer 2

First term:101
Last term:199
U can see that these are in A.P
So
S99=99/2(101+199)
S99=99/2(300)
S99=99 X 150
S99=14850
Hope it helped u!