When the same constant is added to the numbers $60,$ $120,$ and $140,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?
Let that constant be k.
\((120 + k)^2 = (60 + k)(140 + k)\\ k^2 + 240k + 14400 = k^2 + 200k + 8400\\ 40k + 6000 = 0\\ k = -150\)
Common ratio = \(\dfrac{120 + k}{60 + k} = \dfrac13\)
+729 
