Something Hard, But Not Actually?

Take it as an AP... and by using the summation formula, We can get, 104950.

Ʃa = n(a1 + an)/2,

Where n = 100, a1 = 1000,

and an = 1000 + (100-1) = 1099

l=a+(n-1)d

=1000+(100-1)1 = 1099

s= n/2(a+l)

=50(1000+1099)

=104950

sum of n terms of an a.p Sn=n/2{2a+(n-1)d} in this question a=1000 n=100 d=1

Sn=n/2(a+n)

its an AP wit 100 terms starting with 1000 and ending with 1099 and the formula for this is n/2(1st term + last term)