Algebra question -- Solving 2 simultaneous equations...

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It is a bit of a long question about series, But what is important is this...

a+ 2c = 3b

(3b-2)^2 = 2a * (4c-2)

It is asking for the ratio c to a

It have tried a lot of ways. I always end up with this. (Note maybe I not noticing a mistake, Can just anyone confirm that this is solvable? )

a^2 - 4 a c + 4 = 8 c - 4 c^2
 

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  • #2
Math_QED
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It is a bit of a long question about series, But what is important is this...

a+ 2c = 3b

(3b-2)^2 = 2a * (4c-2)

It is asking for the ratio c to a

It have tried a lot of ways. I always end up with this. (Note maybe I not noticing a mistake, Can just anyone confirm that this is solvable? )

a^2 - 4 a c + 4 = 8 c - 4 c^2
You can solve this for a! I got the same expression as you. You can write it as a^2 - 4ac + 4c^2 = 8c - 4.
Now look at the a^2 - 4ac + 4c^2. You should notice something. Solve for a then and you can express c/a
 
  • #3
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You can solve this for a! I got the same expression as you. You can write it as a^2 - 4ac + 4c^2 = 8c - 4.
Now look at the a^2 - 4ac + 4c^2. You should notice something. Solve for a then and you can express c/a
a = 2 ( +or- sqrt( c-1) + c)

Which doesn't make sense in my view, Because the answer is 5 : 2 ( c to a). Did you get the value for c:a as a number or with variables?

Here is the original question:
2a, 3b , 4c is a Arithmetic progression
2a, 3b-2, 4c-2 is a geometric sequence

Find c : a
 
  • #4
Math_QED
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a = 2 ( +or- sqrt( c-1) + c)

Which doesn't make sense in my view, Because the answer is 5 : 2 ( c to a). Did you get the value for c:a as a number or with variables?

Here is the original question:
2a, 3b , 4c is a Arithmetic progression
2a, 3b-2, 4c-2 is a geometric sequence

Find c : a
It might have been useful to include the original question. I don't have time right now. I will look at it tomorrow.
 
  • #5
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This must involve some fancy maneuvers, if it is possible at all. I was able to reduce the single equation we get after eliminating ##b## to the forms:

##(a-2c)^2 = 8c-4##

or

##(2c-2)^2 = 4ac-a^2##

I don't see any way to get ##\frac{c}{a}## out of either of these.
 
  • #6
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Here is the original question:
2a, 3b , 4c is a Arithmetic progression
2a, 3b-2, 4c-2 is a geometric sequence

Find c : a

I wonder if you have left some information out of the problem statement because I get two solutions by brute force search:

## [a,b,c] = [1/4,1/2,5/8] ##
## [a,b,c] = [1,2,5/2] ##

Both of these give ##c/a = 5/2## and both sets fit the given progressions. Additionally, these sets fit the progressions but yields the wrong ##c/a## ratio:

## [a,b,c] = [4,2,1] ##
## [a,b,c] = [16,14,13] ##

These observations lead me to think there is more to the story than was given here. Is there something missing?
 
  • #7
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I wonder if you have left some information out of the problem statement because I get two solutions by brute force search:

## [a,b,c] = [1/4,1/2,5/8] ##
## [a,b,c] = [1,2,5/2] ##

Both of these give ##c/a = 5/2## and both sets fit the given progressions. Additionally, these sets fit the progressions but yields the wrong ##c/a## ratio:

## [a,b,c] = [4,2,1] ##
## [a,b,c] = [16,14,13] ##

These observations lead me to think there is more to the story than was given here. Is there something missing?
Apparently, When the teacher gave me the question the he missed some variables in the geometric sequences. Sorry for inconvenience
 

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