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Geometric sequence, determining the value of the first term

  1. Dec 28, 2006 #1
    question - if the 3rd and 9th term of a geometric series with a positive common ratio are -3 and -192 respectively, determine the value fo the first term, a.

    I kno we using
    a_n=a_1r^n-1


    From that i got this :

    -3 = a_1r^2
    -192 = a_1r^8

    But I dont kno how I can solve for r or a_1... I also don't know if I even did it right
     
  2. jcsd
  3. Dec 28, 2006 #2

    Hurkyl

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    Well, you have two equations in two unknowns, don't you? How do you generally solve such things?
     
  4. Dec 28, 2006 #3
    thats what im wondering.. I dont think the way your thinking of works.. I forgot how to figure out the common ratio thats all i really need to know
     
  5. Dec 28, 2006 #4
    You need to solve the equations you wrote down to get r=

    -3 = a_1r^2
    -192 = a_1r^8
     
  6. Dec 28, 2006 #5
    thats what i dont understand, on other questions i got the common ratio easily by using the term before it but these 2 terms r not after each other so i dont know how to figure them out
    to make it more clear its:
    -3=a1r2
    -192=a1r8
     
  7. Dec 28, 2006 #6
    What happens if you divide the second equation by the first?
     
  8. Dec 28, 2006 #7
    64 = r6 ?
    26 = r6
    r = 2?

    anyone kno if this is right?
     
  9. Dec 28, 2006 #8

    Hurkyl

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    Well, if you computed a_1 too, you could check for yourself to see if it's right!
     
  10. Dec 28, 2006 #9
    Ok...

    ifwe know that term 3 is -3,
    a3 = a1*r^3-1
    -3 = a1*2^2
    -3 = 4*a1
    -3/4 = a1

    should be right
     
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