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The sum of the first three terms of a geometric sequence is 54. The sum of the first six terms is 70. Find the common ratio. (All the terms are real numbers.)

 Apr 30, 2022
 #1
avatar+122385 
+2

Let a  be  the first  term

 

We have that

 

a + ar + ar^2   =  54

a ( 1 + r + r^2)  =  54

a =  54 / (1 +r + r^2)    ⇒   (1 + r + r^2)  =   54/a        (1)

 

And

54 + ar^3 + ar^4 + ar^5  =  70

ar^3 / (1 + r + r^2)    = 16  ⇒     (1 + r + r^2)  =  16/ (ar^3)      (2)

 

Equating (1) , (2)

 

54 / a  =  16 / (ar^3)         [  a   ≠  0 ]

 

 

54  =  16/r^3

 

r^3  =  16/54  =  8/ 27

 

r =   (8/27)^(1/3)   =  2 / 3

 

CORRECTION ....THX, CATMG  !!!

 

 

 

 

cool cool cool

 Apr 30, 2022
edited by CPhill  Apr 30, 2022
 #2
avatar
0

that's not the correct answer 

Guest Apr 30, 2022
 #3
avatar+2401 
+1

I think that there may be a small mistake. 

The sum of the first six terms is 70, so

a + ar + ar^2 + ar^3 + ar^4 + ar^5  =  70

54 + ar^3 + ar^4 + ar^5  =  70

ar^3 + ar^4 + ar^5  = 16

I think your method is really smart since by keeping the amount of terms in both expressions constant, you can just divide. :))

 

=^._.^=

catmg  Apr 30, 2022
 #4
avatar+122385 
+1

Thx for pointing out my error, catmg....will make a correction  !!!!

 

cool cool cool

CPhill  Apr 30, 2022
 #5
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0

CPhill, you really messed up here.

Guest Apr 30, 2022
 #6
avatar+122385 
0

Yep....it happens  when I'm trying to  work too  quickly.....!!!!

 

Hopefully, I've repaired the damage  sad sad sad

 

 

cool cool cool

CPhill  Apr 30, 2022

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