+0

# Algebra

0
2
3
+56

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?

Jul 24, 2024

#1
+1651
+1

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! :)

Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024
#2
+37001
+1

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

ElectricPavlov  Jul 24, 2024
#3
+1651
+1

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

NotThatSmart  Jul 24, 2024
edited by NotThatSmart  Jul 24, 2024