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plohrr

  • 6 months ago

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

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  1. plohrr
    • 6 months ago
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    @Hero @ganeshie8

  2. plohrr
    • 6 months ago
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    @SolomonZelman @jim_thompson5910

  3. jim_thompson5910
    • 6 months ago
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    5r = -10 r = ???

  4. plohrr
    • 6 months ago
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    -2

  5. jim_thompson5910
    • 6 months ago
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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

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

  7. myininaya
    • 6 months ago
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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)

  8. plohrr
    • 6 months ago
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    Ok, so how do i set up this equation

  9. myininaya
    • 6 months ago
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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

  10. plohrr
    • 6 months ago
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    5n = -10^n-1 ?

  11. myininaya
    • 6 months ago
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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

  12. plohrr
    • 6 months ago
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    5_n = 5 x -2^n-1

  13. myininaya
    • 6 months ago
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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

  14. plohrr
    • 6 months ago
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    So how do i set up the equation ? im confused

  15. plohrr
    • 6 months ago
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    Does the 5 not go where the n is

  16. jim_thompson5910
    • 6 months ago
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    5 goes where 'a' is but \(\Large a_n\) is NOT the same as just 'a'

  17. jim_thompson5910
    • 6 months ago
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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

  18. plohrr
    • 6 months ago
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    Ok

  19. jim_thompson5910
    • 6 months ago
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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 ---> \(\Large a_n = a_{17}\)

  20. plohrr
    • 6 months ago
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    Ok, so you leave the an as it is

  21. myininaya
    • 6 months ago
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    Yep it is a formula for the nth term in the sequence

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

  23. myininaya
    • 6 months ago
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    how do you get -15 for r?

  24. myininaya
    • 6 months ago
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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

  25. myininaya
    • 6 months ago
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    also this was already determined way above when joe was working with you

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