sequences

Let the constant be $c$, and let the common difference be $d$.

We have the equation from the first and second terms: $d(60 + c) = 100 + c$, and likewise, we have $d(100 + c) = 180 + c$

Solving for $d$ in both equations gives $d = {100 + c \over 60 + c} = {180 + c \over 100+c}$

Cross multiplying gives us: $c^2 + 200c + 10000 = c^2+240c+10800$

Solving, we find $c =-20$, meaning the series is 40, 80, 160. The common ratio is $8 0\div 40 = \color{brown}\boxed2$