But that's huge!

Notice that for a polynomial expansion of the form:

$(x-a_1)(x-a_2)(x-a_3)\ldots(x-a_n) = x^n - (a_1 + a_2 + a_3 + \ldots + a_n)x^{n-1} + \ldots$

The coefficient of the term with the second highest power will be $-(a_1 + a_2 + a_3 + \ldots + a_n)$

Don't believe? Try it with a few smaller polynomials.

$(x-1)(x-2) = x^2 \color{#3D99F6}{- x - 2x} + 2 = x^2 \color{#3D99F6}{-(1+2)x} + 2\\ (x-1)(x-2)(x-3) = (x^2 - x - 2x + 2)(x - 3) = x^3 \color{#3D99F6}{ - x^2 - 2x^2} + 2x \color{#3D99F6}{- 3x^2} + 3x + 6x - 6 = x^2 \color{#3D99F6}{- (1+2+3)x^2} + 11x - 6$

Therefore,

$(x-1)(x-2)(x-3) \ldots (x-100)\\ = x^{100} \color{#3D99F6}{-(1+2+3+ \ldots + 100)x^{99}} + \ldots$

The coefficient of $x^{99}$

$=-(1+2+3+\ldots+100)\\ =-\left(\displaystyle \sum_{n=1}^{100}n \right)\\ =-\left(\dfrac{100(101)}{2}\right)\\ =-(50)(101)\\ =\boxed{-5050}$

We know that the coefficient of x^99 is just the sum of the roots. So, -b/a=1+2+3+...+100. The coefficient of x^100 is 1 so a is 1. Hence b=-5050