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?

tomtom Jun 15, 2024

#1**+1 **

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! :)

NotThatSmart Jun 15, 2024

#1**+1 **

Best Answer

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! :)

NotThatSmart Jun 15, 2024