The first four terms of the series probably refer to the first four partial sums.
[tex]\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4[/tex]
[tex]\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3[/tex]
[tex]\displaystyle\sum_{n=1}^3-4\left(\frac13\right)^{n-1}=-\frac{52}9[/tex]
[tex]\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}[/tex]
which you can compute either by adding one term at a time, or using the well-known formula,
[tex]\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}[/tex]
(I can provide a link to a derivation I gave in a nearly identical question in the comments)
The series converges as [tex]k\to\infty[/tex] if and only if [tex]|r|<1[/tex], which is certainly the case here, since [tex]-1<\dfrac13<1[/tex].
Extrapolating from the formula above, the sum of the convergent series is
[tex]\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}[/tex]
so the sum for this series is [tex]\dfrac{-4}{1-\frac13}=-6[/tex].
