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The fifth term of a geometric sequence of positive numbers is 11 and the eleventh term is 5. What is the eighth term of the sequence? Express your answer in simplest radical form.

 

Please HElp Thanks!!!

 Mar 6, 2019
 #1
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Let the first term =  a1

 

5th term =     a1*r^4 =  11   ⇒    a1 =  11/r^4    (1)

11th term   = a1*r^10 = 5       (2)

 

Subbing (1) into (2) we have

 

(11/r^4) * r^10  = 5

 

11r^6  =  5

 

r^6  = 5/11       take the 6th root of both sides

 

r = ( 5/11)^(1/6)

 

a1 = 11/r^4 =  11/ [(5/11)^(1/6)]^4  =    11/ (5/11)^(4/6)

 

So...the 8th term is

 

a1*r^7  =

 

 ( 11/ (5/11)^(4/6) ) * [(5/11)^(1/6)]^7    =

 

11 [ (5/11]^(7/6)] / (5/11)^(4/6)  =

 

11 * (5/11)^(3/6)  =

 

11 * (5/11)^(1/2) =

 

11*√5 / √11 =

 

√11* √5 =

 

√55

 

 

cool cool cool

 Mar 6, 2019

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