+657 When the same constant is added to the numbers $60,$ $120,$ and $160,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?
+897 This problem isn't too bad to solve!
First off, we need an equation to figrue out the 3 new numbers we achieve. Let's let n be the constant we add.
Since all terms of a geomtetric sequence have equal fractions, we can write the equation
\(\frac{120+n}{60+n}=\frac{160+n}{120+n}\\ (120+n)^2=(60+n)(160+n)\\ 14400+n^2+240n=9600+n^2+220n\\ 4800=-20n\\ n=-240\)
I know the constant is now negative, but you do get the numbers -180, -120, and -80, which do in fact form common ratios.
-180/-120 = 3/2. So 3/2 is the answer.
I'm not sure if this was the exact answer you were looking for, but it does work
Thanks! :)
