+126 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?
+1786 Let's write an equation for this problem.
Since they form a geometric series, let's let n be the number that is added.
We have the equation
\(\frac{100+n}{60+n}=\frac{160+n}{100+n}\\ (100+n)^2=(60+n)(160+n)\\ 10000+n^2+200n=9600+n^2+220n\\ 400=20n\\ n=20\\\)
Thus, adding 20 to each term, we get the sequence
80, 140, 180.
Since
\(120/80 = 3/2 \\ 180/120=3/2\)
The common ratio is just 3/2.
Thanks! :)
+37044 Hey, NotThatSmart has the right method, but accidentally used 100 instead of 120 as the second term :
(120+n)/(60+n) = (160+n)/(120+n) cross multiply to get
n^2 + 240n + 14400 = n^2 +220n + 9600
20n = -4800
n = -240 would be the constant to add to the numbers ....
SO the geometric sequence would be
-180 -120 -80 and the common ratio would be -120/-180 = 2/3
+1786 Oh yeah..nice catch EP. I don't know what my brain was thinking.
It was 9 in the morning, not sure if that's an excuse...LOL
Nice work EP. Sorry for the mistake.
I like how I just switched between the two numbers randomly...
~NTS
