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sasogeek

is it okay to do this...? and did i get it correct anyway? the third term of an exponential sequence is 12 and the 9th term is 768. find the values of the common ratio. \[\large ar^2=12 \]\[\large ar^8=768\] \[\large \frac{ar^8}{ar^2}=\frac{768}{12}\] \[\large r^6=64 \] \[\large r=2 \ , \ r=-2\] I forgot the formulas and decided to go with this but i don't know if it is correct or even if it's acceptable... can anyone confirm for me please?

  • one year ago
  • one year ago

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  1. sasogeek
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    truth is i've even forgotten if this is exactly how to do it lol. i literally have no memory of the classes i took for this topic

    • one year ago
  2. .Sam.
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    yeah you can do that

    • one year ago
  3. ParthKohli
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    You are perfectly correct, math can be done in infinite ways :)

    • one year ago
  4. Mimi_x3
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    it looks like a geometric progression.. i dont know what's an exponential sequence.. http://en.wikipedia.org/wiki/Exponential_sheaf_sequence

    • one year ago
  5. .Sam.
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    \[a_3=12\] \[a_9=768\] ------------- \[ar^2=12\] \[ar^8=768\] \[\frac{12}{r^2}r^8=768\] \[r^6=64\] \[r= \pm 2\]

    • one year ago
  6. sasogeek
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    same thing :) different name

    • one year ago
  7. .Sam.
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    GP formula \[\Huge T_n=a_1r^{n-1}\]

    • one year ago
  8. sasogeek
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    thanks :)

    • one year ago
  9. .Sam.
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    np

    • one year ago
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