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Geometric Progressions Help

  1. Apr 28, 2004 #1
    I am having toruble with my geometric progressions, in that i ahv ebeen given a question where i am given the 7th and 26th terms of a GP. I am required to find the ratio however, which i could do if i had the first term. Usually i can do this as they only give me gps that are one term apart, and i would divide the top by bottom (say Term6 = 3 and term7 = 4) and woudl end up with term1 = 3/4. How can i do this if the terms are as far apart as they are?

    Welcoming any responses here :smile:
  2. jcsd
  3. Apr 28, 2004 #2


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    i think you have too many variables such as a1 and n (the number of terms) that are unknown at least one of them are needed to solve for the quotinent.
  4. Apr 28, 2004 #3


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    You know that in a geometric progression, the next term's ratio with the previous is a constant; let's call it x; that is GP(n+1)/GP(n)=x.
    But then we must have: GP(n+2)/GP(n)=(GP(n+2)/GP(n+1))*GP(n+1)/GP(n)=x^(2).
    Did that help?
  5. Apr 29, 2004 #4
    Thanks for the help, ive got it down pat now :) hope im ready for maths test tomorrow :wink:
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