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plohrr Group Title

what are the explicit equation and domain for a geometric sequence with a first term of 5 and a second term of -10

  • one month ago
  • one month ago

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  1. plohrr Group Title
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    @Hero @ganeshie8

    • one month ago
  2. plohrr Group Title
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    @SolomonZelman @jim_thompson5910

    • one month ago
  3. jim_thompson5910 Group Title
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    5r = -10 r = ???

    • one month ago
  4. plohrr Group Title
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    -2

    • one month ago
  5. jim_thompson5910 Group Title
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    First Term: a = 5 Common Ratio: r = -2 Plug those into \[\Large a_{n} = a*(r)^{n-1}\] to get the general nth term formula

    • one month ago
  6. plohrr Group Title
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    5n = 5 x (-2) n-1 5n = -10 n-1 What do i do from here

    • one month ago
  7. myininaya Group Title
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    no a_n symbolizes the n term of the sequence a_n and a are different values also we don't have (ar)^(n-1) we just have that the r is to the (n-1)

    • one month ago
  8. plohrr Group Title
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    Ok, so how do i set up this equation

    • one month ago
  9. myininaya Group Title
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    \[a_n=a \cdot r^{n-1}\] just replace the a with 5 and replace the r with -2 like jim said

    • one month ago
  10. plohrr Group Title
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    5n = -10^n-1 ?

    • one month ago
  11. myininaya Group Title
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    again a_n and a are different values a_n is a variable by itself you can't multiply a and r because r has an exponent

    • one month ago
  12. plohrr Group Title
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    5_n = 5 x -2^n-1

    • one month ago
  13. myininaya Group Title
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    you cannot replace a_n with 5_n this makes no sense let me show you how this formula is come up with so maybe you understand better {a_n} is a sequence of numbers those numbers are a_1=5 a_2=5(-2)=-10 a_3=5(-2)(-2)=5(-2)^2=20 a_4=5(-2)(-2)(-2)=5(-2)^3=-40 a_5=5(-2)(-2)(-2)(-2)=5(-2)^4=80 ... therefore a_n=5(-2)(-2)(-2)(-2)(-2)(-2)...(-2)= (by the way that is a (n-1) amount of (-2)) I put a (n-1) amount of (-2) because looking at a_1 we have (-2)^0=(-2)^(1-1) looking at a_2 we have (-2)^1=(-2)^(2-1) looking at a_3 we have (-2)^2=(-2)^(3-1) it is always one less than where the number is in the sequence

    • one month ago
  14. plohrr Group Title
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    So how do i set up the equation ? im confused

    • one month ago
  15. plohrr Group Title
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    Does the 5 not go where the n is

    • one month ago
  16. jim_thompson5910 Group Title
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    5 goes where 'a' is but \(\Large a_n\) is NOT the same as just 'a'

    • one month ago
  17. jim_thompson5910 Group Title
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    If it confuses you, think of it as \[\Large T = a*(r)^{n-1}\] T = nth term a = first term r = common ratio n = positive whole number (used to identify which term you're dealing with) example: n = 3 ---> 3rd term

    • one month ago
  18. plohrr Group Title
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    Ok

    • one month ago
  19. jim_thompson5910 Group Title
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    The notation \(\Large a_n\) is used for sequences because the 'n' changes to a positive whole number to indicate a certain term example: 17th term ---> n = 17 ---> \(\Large a_n = a_{17}\)

    • one month ago
  20. plohrr Group Title
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    Ok, so you leave the an as it is

    • one month ago
  21. myininaya Group Title
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    Yep it is a formula for the nth term in the sequence

    • one month ago
  22. plohrr Group Title
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    So it is an = 5 (-15)^n-1; all integers where n less than or equal to 1

    • one month ago
  23. myininaya Group Title
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    how do you get -15 for r?

    • one month ago
  24. myininaya Group Title
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    r symbolizes the geometric ratio to find r all you have to do is evaluate a_2/a_1 or take any two consecutive numbers in the sequence and put the 2 nd term of those numbers over the 1st term of those numbers

    • one month ago
  25. myininaya Group Title
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    also this was already determined way above when joe was working with you

    • one month ago
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