We can write:

a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2

Hence we have: na + (n-1)n/2 = 100, which we can rewite as: n^{2} + (2a-1)n - 200 = 0

Solve this for n in terms of a: \(n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}\) (the other solution results in negative values for n).

Let a run from 1 to 100 to find the only integer solutions are: (a,n) = (9,8) and (18,5) as found by Chris and Guest.

(There is also a = 100 and n = 1, but the question requires n>=2).

(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)