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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?

 Jun 15, 2024

Best Answer 

 #1
avatar+1280 
+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! :)

 Jun 15, 2024
 #1
avatar+1280 
+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

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