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t_1 (first term): p^2/2

 

r (common ratio): p/2

 

t_n (nth term): p^9/256

 Nov 6, 2017
 #1
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We need to solve this :

 

(p^2 / 2 ) * (p/2)^(n -1)  = p^9 / 256    where n is the nth term

 

Multiply through by  2 / p^2

 

(p / 2)^(n -1)  =  p^7 / 128

 

Note   p^7 / 128  =   p^7 / 2^7  =  ( p / 2)^7

 

So......

 

(p / 2) ^(n - 1)  =  (p /2 )^7

 

Since the bases are the same, solve for the exponents

 

n - 1  = 7      add 1 to both sides

 

n  =  8   terms

 

 

cool cool cool

 Nov 6, 2017

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