If $a$ is the first term and $-d$ is common difference then $n^{\textrm{th}}$ term is $a+(n-1)(-d)$.
Now if in this AP if we are asked what will be the first negative term then how does $a+(n-1)(-d)<0$ give the answer?
I mean why is it giving the first negative term why not any other negative term?
It doesn't give the first negative term. It gives all values of $n$ which produce negative terms. If you continue, you have
$$a + (n-1)(-d)< 0$$ $$(n-1)(-d)< -a$$ $$n-1 > \frac{-a}{-d}=\frac ad\tag{assuming $d>0$, so $-d<0$}$$ $$n > 1 + \frac ad$$ Every positive integer greater than $1+\frac ad$ will produce a negative term. The first such integer will produce the first negative term, and all terms thereafter will also be negative.
For example, if $a=1$ and $d=2$, then you get $n>1+1/2=1.5$, so the positive integers satisfying $n>1.5$ are $2,3,4,5,\cdots$ and the first of these is $2$; so the second term is the first negative term.
You can verify this by observing that the corresponding progression is $$1, \underbrace{-1}_{2^{\textrm{nd}}\textrm{ term}}, -3, -5, -7, \ldots$$