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?

Pythagorearn May 28, 2024

#1**+1 **

Not too bad of a problem!

Let's let n be the constant we add to the 3 numbers. Lets try to figure out the 3 new numbers first.

Since all terms form a geometric sequence, we can write

\(\frac{120+n}{60+n}=\frac{160+n}{120+n}\\ (120+n)^2=(60+n)(160+n)\\ 14400+n^2+240n=9600+n^2+220n\\ 4800=-20n\\ n=-240\)

Now, I know the n is negative, when if we apply it, we dot get -180, -120, and -80, which do form a geometric series.

-180/-120 = 3/2, so 3/2 is the answer.

I'm not sure if this is the correct answer, but it does work.

Thanks! :)

NotThatSmart May 28, 2024