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If the second term of a geometric sequence of real numbers is -2 and the fifth term is 16 then what is the fourteenth term?

 Apr 15, 2017
 #1
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We have that

 

-2  = a1*(r)^(2-1)  →  -2 = a1*(r)         (1)

 

And 

 

16  =  a1* (r)^(5 -1)  →  16  = a1*(r)^(4)     (2)

 

Rearrange (1)  as   -2/r   =  a1      sub this into (2)

 

16  =  (-2/r) * (r)^4      simplify

 

16  = (-2) *(r)^3       divide both sides by  -2

 

-8  =  (r)^3        take the cube root of each side and   -2  = r

 

Using  -2/r  =   a1     →   -2/-2   =  a1   =  1

 

So....the foruteenth term is

 

1(-2)^(14 - 1)  =  1(-2)^13   =   - 8192

 

 

 

cool cool cool

 Apr 15, 2017

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