Let the numbers be
$a-5 d, a-3 d, a-d, a +d, a+3 d, a+5 d$
$\therefore a-5 d+a-3 d +a -d+ a+ d +a$
$+3 d+a+5 d=3$
$(\because$ Sum $=3)$
$\Rightarrow 6 a=3 \Rightarrow a=\frac{1}{2}$
Also, given $T_{1}=4 T_{3}$, where $T_{1}, T_{3}$ are respectively, first and third terms of AP.
$\Rightarrow a-5 d=4(a-d)$
$\Rightarrow d=-3 a=-\frac{3}{2}$
$\therefore $ The fifth term
$a+3 d=\frac{1}{2}+3\left(-\frac{3}{2}\right)=\frac{1}{2}-\frac{9}{2}=-4$