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October 2nd, 2019, 09:47 AM   #1
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Geometric sequence

How can I find x so that a+x, b+x, c+x is a geometric sequence?

Last edited by skipjack; October 2nd, 2019 at 07:52 PM.
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October 2nd, 2019, 10:05 AM   #2
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You need (a + x)(c + x) = (b + x)².
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October 2nd, 2019, 10:17 AM   #3
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Now just set a,b,c the way you want and find their values.

Note: If (p,q,r) is a geometric sequence, (r,q,p) is also a geometric sequence.

Last edited by skipjack; October 2nd, 2019 at 07:51 PM.
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October 2nd, 2019, 04:37 PM   #4
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Found a solution

Quote:
Originally Posted by skipjack View Post
You need (a + x)(c + x) = (b + x)².
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October 2nd, 2019, 04:38 PM   #5
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Funny but true

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Originally Posted by idontknow View Post
Now just set a,b,c the way you want and find their values .

Note: If {p,q,r} is a geometric sequence then {r,q,p} is also a geometric sequence .
You're actually correct

1,3,9 and 9,3,1 are both geometric sequences.
First one has a ratio of 3, the second one has a ratio of 1/3
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October 2nd, 2019, 07:54 PM   #6
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What if no such x exists?
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October 3rd, 2019, 02:54 AM   #7
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I think that will happen if $\displaystyle b$ is the arithmetic mean of $\displaystyle a$ and $\displaystyle c$.
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October 3rd, 2019, 03:01 AM   #8
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Quote:
Originally Posted by mrtwhs View Post
I think that will happen if $\displaystyle b$ is the arithmetic mean of $\displaystyle a$ and $\displaystyle c$.
for x=1 : {1+a,1+b,1+c}. Now set a=1,b=3,c=7 and we have a geometric sequence {2,4,8} where the ratio is $\displaystyle q=2$.
Also {8,4,2} where the ratio is $\displaystyle q=1/2$.
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