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Suppose x-2, x and x+6 where are sequential terms in a geometric sequence

  1. May 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose x-2,x and x+6, where x is an integer, are consecutive terms in a geometric sequence S
    Determine x

    2. Relevant equations
    r=x/(x-2)=(x+6)/x

    3. The attempt at a solution
    x/(x-2)=(x+6)/x cross multiply
    x(x)=(x-2)(x+6)
    x^2=x^2+4x-12
    x^2-x^2+4x-12/4=0
    4x=12
    x=3
     
    Last edited: May 7, 2015
  2. jcsd
  3. May 7, 2015 #2

    SteamKing

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    You might want to double check your algebra when solving for x.
     
  4. May 7, 2015 #3
    @SteamKing I have corrected the -to+, am I doing the right thing?
    I am quite rusty with math, I have started a degree towards civil engineering now at the age of 33, I last practiced math in 2000.

    Thank you for your input.
    Jaco
     
  5. May 7, 2015 #4

    Ray Vickson

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    Now is the time to learn to break bad habits: Never write something like x/x-2, because the means ##\frac{x}{x} - 2 = 1 - 2 = -1## when read using standard parsing rules for mathematical expressions. You want ##\frac{x}{x-2}##, so you need parentheses, like this: x/(x-2). Similarly for your other expression x+6/x, which means ##x + \frac{6}{x}## as you have written it. You should write (x+6)/x.
     
  6. May 7, 2015 #5
    Thank you for pointing this out Ray,
    I did pick up the error on the following statements before posting but have corrected these swell.
    Does my answer make sense?

    Thank you again.
     
  7. May 7, 2015 #6

    SammyS

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    What do you think?

    Can you show that it does indeed make sense?
     
  8. May 7, 2015 #7

    HallsofIvy

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    Apparently you have edited your original solution. You now have x= 3 in which case x- 2= 1, x= 3, and x+ 6= 9 so the sequence is 1, 3, 9 which is the same as [itex]3^0, 3^1, 3^2[/itex]. Yes, that is a geometric sequence. With you original answer, which was apparently x= -3, x- 2= -5, x= -3, x+ 6= 3. The sequence -5, -3, 3 is NOT a geometric sequence.
     
    Last edited: May 7, 2015
  9. May 7, 2015 #8

    SteamKing

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    Yes, your calculation is now correct.

    A word of advice here. Always check your calculations. It's very easy for arithmetic mistakes to creep into a calculation. Making mistakes like this will be important to avoid later on in studying for your degree. Silly mistakes can cost you points on exams and assignments.
     
  10. May 7, 2015 #9
    Thank you everyone,
    I appreciate the help, distance learning has its downfall with math.
    Have a great day.
     
  11. May 7, 2015 #10
    a follow up question to this:
    Suppose (x-2) is the 4th term, determine the first:
    An=A1*34-1
    (x-2)=A1*33
    (x-2)=A*27
    (x-2)=27A
    A=(x-2)/(27)
    A=1/27 (x-2) was determined in the previous answer as 1
    =3-3
     
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