If $x$ is positive, the first negative term in the expansion of $(1+x{)}^{27/5}$ is:

In the expansion $(1+x{)}^n$, the general term is $T_{r+1}=\frac{n(n-1)(n-2)\dots (n-r+1)}{r!}{x}^r$.For $n=\frac{27}{5}$, the term becomes negative when one of the factors in the numerator becomes negative.The factor $(n-k)$ becomes negative when $k>n$.Here $n=5.4$.Factors are $n,n-1,n-2,n-3,n-4,n-5,n-6$.$n-6=5.4-6=-0.6$ is the first negative factor.This factor occurs in the 8th term (where $r=7$ and the last factor is $n-7+1=n-6$).Thus, the 8th term is the first negative term.