+1911 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?
+1908 Let's set variables in order to solve this problems.
Let's let c be the constant added to each number.
We get \(60+c, 120+c, 160+c\)
Since all the terms form an geometric series, we can write the formula
\(\frac{60+c}{120+c} = \frac{120+c}{160+c}\)
Now, we simplfy solve for c.
\((c+160)(60+c)=(c+120)(120+c)\)
Distributing everything, combining all like terms, and moving everything to one side, we get
\(-20c-4800=0\)
\(c=-240\)
We plug this back in, and we get the series \(-180, -120, -80\)
This has a common ratio of \(180/120=3/2\)
So 3/2 is our answer,
Thanks! :)
+1908 Let's set variables in order to solve this problems.
Let's let c be the constant added to each number.
We get \(60+c, 120+c, 160+c\)
Since all the terms form an geometric series, we can write the formula
\(\frac{60+c}{120+c} = \frac{120+c}{160+c}\)
Now, we simplfy solve for c.
\((c+160)(60+c)=(c+120)(120+c)\)
Distributing everything, combining all like terms, and moving everything to one side, we get
\(-20c-4800=0\)
\(c=-240\)
We plug this back in, and we get the series \(-180, -120, -80\)
This has a common ratio of \(180/120=3/2\)
So 3/2 is our answer,
Thanks! :)
