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anonymous

  • one year ago

HELP PRECALC GIVING MEDALS AND BECOMING A FAN 1.)Find an explicit rule for the nth term of the sequence. -4, -8, -16, -32 an = -4 2^(n - 1) an = 2 -4^(n + 1) an = 2 -4^n an = -4 2^n 2.) Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -12 and 768, respectively. an = 3 (-4)^(n + 1) an = 3 4^(n - 1) an = 3 (-4)^(n - 1) an = 3 4^n

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    you can always check which ones work, but each term in negative, so the number out fron must be negative

  3. anonymous
    • one year ago
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    For Question one this is answer: an = -4 • 2n - 1

  4. misty1212
    • one year ago
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    if \(n=1\) you should get \(-4\) and i think only \[-4\times 2^{n-1}\] works

  5. anonymous
    • one year ago
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    @misty1212 so is an = -4 • 2^(n - 1) correct?

  6. misty1212
    • one year ago
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    yes

  7. anonymous
    • one year ago
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    thank you both @misty1212 for the second question I know it isn't an = 3 4^(n - 1) but i hv no idea what it is

  8. anonymous
    • one year ago
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    @freckles

  9. misty1212
    • one year ago
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    you are supposed to get something that looks like \[a_n=a_0\times r^{n-1}\]

  10. misty1212
    • one year ago
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    \[a_2=-12\\ a_5=768\]

  11. anonymous
    • one year ago
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    is it A? @misty1212

  12. misty1212
    • one year ago
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    divide and get \[\frac{a_5}{a_2}=\frac{768}{-12}=-64\]

  13. misty1212
    • one year ago
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    idk i can't eyeball it, we have to do it

  14. misty1212
    • one year ago
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    that means \[r^3=-64\] so \[r=-4\]

  15. misty1212
    • one year ago
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    then go with C since it is the one that looks like \[a_0r^{n-1}\] in this case \(r=-4\) and \(a_0=3\) since \[3\times (-4)^{2-1}=3\times -4=-12\]

  16. anonymous
    • one year ago
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    thanks @misty1212

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