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In a geometric sequence u1=125 and u6 =1/25

(A) find the value of r (the common ratio)

(B) find the largest value of n for which Sn < 156.22

(C)explain why there is no value of n for which Sn > 160

 

Can someone please help me figure this out?

 Mar 2, 2021
 #1
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u1  =   125

u6  = 1/25

                           

A)   

 

125 r^5  =  1/25

 

r^5  =  (1/25) * (1/125)

 

r^5  =  (1/ 5^2) * (1/5^3)

 

r^5  =  1/ 5^3        take the 5th root

 

r  =  1/5

 

B)   

 

125  ( 1 - (1/5)^n)  / ( 1-1/5)  <  156.22

 

125 ( 1 - (1/5)^n)  / (4/5)   <  156.22

 

156.25 ( 1 - (1/5)^n)  < 156.22

 

1  - (1/5)^n  <   156.22/156.25

 

1 - 156.22/156.25  < (1/5)^n                { 1/5   = .2 }

 

3 / 15625  <   .2^n          take  the log of both sides

 

log  (3/15625)  <   n log (.2)

 

Divide  both sides  by  log (.2)......this results in a negative.....so....reverse the inequality sign

 

log (3/15625)  / log (.2)  >  n

 

5.3  >  n

 

n  = 5

 

 

C)    125  ( 1  - (1/5)^n) /  (4/5)  =   160

 

156.25  ( 1 - (1/5)^n)  =  160             (1)

 

Note  that  ( 1  -(1/5)^n)    CANNOT  be  any  larger  than 1

 

So  ....as large  as  the  left side of  (1)  can be is 156.25   

 

 

cool cool cool

 Mar 3, 2021

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