(a) \[ar=-120,\quad\frac a{1-r}=160\Rightarrow160r(1-r)=-120.\] Hence \[4r^2-4r-3=(2r+1)(2r-3)=0.\] Convergence requires \(|r|<1\), so \(r=-1/2\), \(a=240\).

(b) The first remaining term is the tenth: \[t_{10}=240(-1/2)^9=(-120)(-1/2)^8=-15/32.\] Its infinite tail sum is \[\frac{-15/32}{1+1/2}=-5/16=-0.3125.\]