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# intermediate algebra

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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
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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  !!!   Apr 30, 2022
edited by CPhill  Apr 30, 2022
#2
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Guest Apr 30, 2022
#3
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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
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Thx for pointing out my error, catmg....will make a correction  !!!!   CPhill  Apr 30, 2022
#5
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CPhill, you really messed up here.

Guest Apr 30, 2022
#6
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Yep....it happens  when I'm trying to  work too  quickly.....!!!!

Hopefully, I've repaired the damage      CPhill  Apr 30, 2022