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Suppose w,x,y,z are positive real numbers such that w,x,y,z form an increasing arithmetic sequence, and w,x,z form a geometric sequence.

What is the value of w/z ? The answer must be in fraction, dont divide it.

 

 

Thank you!!!

 May 10, 2022
 #1
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With a bit of trial and error, we can find the sequence to be 2,4,6,8.

 

Can you take it from here?

 May 10, 2022
 #2
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I kind of understandbut could you please deeply elaborate on how to get the answer since I have multiple question like that so if you ahve a method I could find it useful.

 May 10, 2022
 #3
avatar+1810 
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Sorry about that, hopefully, this will make it easier for you!!

 

Let x be the first term and a be a common difference.


We have the arithmetic sequence: \(x, x+a, x+ 2a, x+ 3a\) and the geometric sequence is: \(x, x+a, x+ 3a\)

 

From the geometric sequence, we can derive the equation: \(\large{{{x +a} \over x} ={ {x+3a} \over {x+a}}}\)

 

Basically, this equation states that the common difference between the first 2 terms (x+a and x) is the same as the common difference between the next 2 terms (x+3a and x+a). 

 

 From the geometric series, we know that \({{x+3a}\over x} = 2a\) (value of 3rd term divided by 1st term is 2 times the common difference)

 

When we solve this system, we find that \(x = a = 2\), meaning the series is 2,4,6,8. 

BuilderBoi  May 10, 2022

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