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anonymous

  • one year ago

What is the nth term of the geometric sequence 4, 8, 16, 32, ... ?

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  1. anonymous
    • one year ago
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    A. 2(4)^N-1 B. 4(2)^N-1 C. 2(N)^4 D. 4(N)^2

  2. Vocaloid
    • one year ago
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    formula is (first term)*(common ratio)^(N-1) what is the first term?

  3. anonymous
    • one year ago
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    \[a _{n}\]

  4. Vocaloid
    • one year ago
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    not quite, take a look at your sequence what is the first number in the list?

  5. jdoe0001
    • one year ago
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    ditto

  6. anonymous
    • one year ago
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    A. \[A. a_{n}=2(4)^{n-1} B. a_{n}=4(2)^{n-1} C.a_{n}=2(n)^{4} D. a_{n}=4(n)^{2}\]2(4)^N-1 B. 4(2)^N-1 C. 2(N)^4 D. 4(N)^2

  7. Vocaloid
    • one year ago
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    would you mind answering my question, please? your sequence is 4, 8, 16, 32, ... , right? what is the first number in the sequence?

  8. anonymous
    • one year ago
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    idk

  9. Vocaloid
    • one year ago
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    which number comes first in the list 4, 8, 16, 32, ... ?

  10. anonymous
    • one year ago
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    4

  11. Vocaloid
    • one year ago
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    right! that means our first term is 4 our sequence is 4, 8, 16, 32, ... each consecutive number gets multiplied by 2 to get to the next number, right? our common ratio is 2 so, first term = 4, and common ratio = 2 now, let's look at the formula (first term)*(common ratio)^(N-1) can you tell me which one is the answer now? :)

  12. anonymous
    • one year ago
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    A

  13. Vocaloid
    • one year ago
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    not quite, check carefully (first term)*(common ratio)^(N-1) first term is 4 common ratio is 2

  14. anonymous
    • one year ago
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    B

  15. Vocaloid
    • one year ago
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    much better, good job

  16. anonymous
    • one year ago
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    What is the tenth term of the geometric sequence that has a common ratio of 1/3 and 36 as its fifth term? A. 4/7 B. 27/4 C. 4/81 D. 1/36

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