The coefficient of $x^{49}$ in the product $(x-1)(x-3) \dots(x-99)$ is

Here,

$d=$ difference of terms

$a=1$ st tepms

$n=$ number of terms

we know,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Here we can see that the constant term is starting from 1 and ending at 99 . with the difference of 2 .

we also know

$\begin{array}{l} A_{n}=a+(n-1) d \\ 99=1+(n-1)^{2} \\ \Rightarrow n=50 \end{array}$

Now, when we will multiply the terms starting $(x-1)$ and ending $(x-97)$,

then we will get $x^{49}$.

so,

$\begin{aligned} \text { co-efficient of } x^{49} & =-(1+31 \ldots+99) \\ x^{49} & =-\frac{50}{2} \times(1+99) \\ x^{49} & =-25 \times 100 \\ & =-2500 \end{aligned}$