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Find the second term of a geometric sequence whose third term is 4/3, and whose sixth term is -32/27.

Guest Nov 18, 2018
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Let r be the common ratio

Let the first term  = a

So.....the third term is   ar^2

And the sixth term is  ar^2

 

So.....we have this ratio

 

ar^2          (4/3)   

____  =    _______

ar^5          (-32/27)

 

 

r^2           (4/3)

___  =     ______       and we can invert these fractions and write

r^5          (-32/27)

 

 

r^3   =  (-32/27)  / (4/3)

 

r^3  =  (-32/27) * ( 3/4)   =  (-8) / (9)

 

r =  (-8/9)^(1/3)  =   -2 / (9)^(1/3)   =   - 2 / (3^2)^(1/3)  =  - 2 / (3)^(2/3)

 

So....to find the second term, we have

 

(2nd term) * r  =  (3rd term)          divide both sides by r

 

(2nd term )  =    (3rd term ) / r

 

(2nd term)   =   (4/3)  / [  -2 / (3)^(2/3) ]  =  (4/3) * (3)^(2/3) / -2  = 

 

(4 / -2)  * ( 3^(2/3) / 3 )  =

 

-2 / (3)^1/3 

 

 

cool cool cool

CPhill  Nov 18, 2018

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