Let, $v_1, v_{\mathrm{g}}$ and $v_0$ be the velocities of jet, ejected gases i.e. combustion products and observer on the ground respectively.
Let jet be moving towards right (+ve direction).
$\therefore$ ejected gases will move towards left (-ve direction).
$\therefore$ According to the statement $v_1=500 \mathrm{kmh}^{-1}$
As oberver is at ground i.e. at rest
$$
\therefore v_0=0
$$
Now relative velocity of plane w.r.t. the observer
$$
v_1-v_0=500-0=500 \mathrm{kmh}^{-1}
$$
Relative velocity of the combustion products w.r.t. jet plane $v_g-v_1=-1500 \mathrm{kmh}^{-1}$ (given)
-ve sign indicates that the combustion products move in a direction opposite to that of jet.
$\therefore$ Adding equations (i) and (ii), we get the speed of combustion products w.r.t. observer on the ground i.e.
$$
\left(v_1-v_0\right)+\left(v_g-v_i\right)=v_g-v_0=500+(-1500)
$$
or $v_{g 0}=v_g-v_0=-1000 \mathrm{kmh}^{-1}$
-ve sign shows that relative velocity of the ejected gases w.r.t. observer is towards left i.e. -ve direction i.e. in a direction opposite to the motion of the jet plane.