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marcybaby

  • 3 years ago

Find the general term an for the geometric sequence a1=-3 and a2=6?

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  1. marcybaby
    • 3 years ago
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    \[a_{1}=-3 and a_{2}=6\]

  2. RadEn
    • 3 years ago
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    use the formula : an = a1 r^(n-1) with a1 is given -3 and to get the ratio(r), defined r = a2/a1 = 6/-3 = -2 therefore, an = a1 * r^(n-1) an = -3 * (-2)^(n-1) |dw:1367598379663:dw|

  3. marcybaby
    • 3 years ago
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    so that's the answer? @RadEn I'm confused

  4. nincompoop
    • 3 years ago
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    it is not the answer, rather hint as how to solve the problem.

  5. marcybaby
    • 3 years ago
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    ahhh!! i'm so lost and confused!! @nincompoop

  6. waterineyes
    • 3 years ago
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    Can you find here the common ratio "r" ??

  7. marcybaby
    • 3 years ago
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    -2?

  8. waterineyes
    • 3 years ago
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    Yep..

  9. waterineyes
    • 3 years ago
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    So, general term can be found as: \[\large \color{green}{a_n = a \cdot r^{n-1}}\]

  10. waterineyes
    • 3 years ago
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    a = -3 and r = -2.. Plug in and find..

  11. marcybaby
    • 3 years ago
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    6?

  12. waterineyes
    • 3 years ago
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    How?

  13. waterineyes
    • 3 years ago
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    Use that formula,

  14. marcybaby
    • 3 years ago
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    -3(-2)^n-1?

  15. waterineyes
    • 3 years ago
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    Yep, this is the general term you require...

  16. marcybaby
    • 3 years ago
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    okay, but I am stuck here. I dont what is the next step

  17. marcybaby
    • 3 years ago
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    i get 6^n-1?

  18. waterineyes
    • 3 years ago
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    Hey, this is the required answer, you can't solve it further..

  19. marcybaby
    • 3 years ago
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    so this cant get be simplified any more?

  20. waterineyes
    • 3 years ago
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    No..

  21. waterineyes
    • 3 years ago
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    \[\large a_n = -3\cdot (-2)^{n-1}\]

  22. marcybaby
    • 3 years ago
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    okay... how come tho? 6^n-1 looks like an ideal answer to me.

  23. waterineyes
    • 3 years ago
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    You cannot multiply \(-3\) with \((-2)^{n-1}\)

  24. marcybaby
    • 3 years ago
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    ohhhhhhhhhhh!!!!!!!!!! duh!! now i get it! my brain hurts been studing all week.

  25. waterineyes
    • 3 years ago
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    It is okay..

  26. marcybaby
    • 3 years ago
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    thank you for your time and patience

  27. waterineyes
    • 3 years ago
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    You are welcome dear..

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