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 October 2nd, 2019, 09:47 AM #1 Newbie   Joined: Oct 2019 From: San diego Posts: 3 Thanks: 0 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.
 October 2nd, 2019, 10:05 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 You need (a + x)(c + x) = (b + x)². Thanks from idontknow
 October 2nd, 2019, 10:17 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 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.
October 2nd, 2019, 04:37 PM   #4
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Found a solution

Quote:
 Originally Posted by skipjack You need (a + x)(c + x) = (b + x)².
Geometric Sequence – Math tutor in San Diego

October 2nd, 2019, 04:38 PM   #5
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Funny but true

Quote:
 Originally Posted by idontknow 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

 October 2nd, 2019, 07:54 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,035 Thanks: 2271 What if no such x exists?
 October 3rd, 2019, 02:54 AM #7 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 I think that will happen if $\displaystyle b$ is the arithmetic mean of $\displaystyle a$ and $\displaystyle c$. Thanks from idontknow
October 3rd, 2019, 03:01 AM   #8
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Quote:
 Originally Posted by mrtwhs 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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