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anonymous

  • one year ago

Identify the 12th term of a geometric sequence where a1 = 8 and a6 = -8,192. 134,217,728 33,554,432 -33,554,432 -134,217,728

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  1. Michele_Laino
    • one year ago
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    the general formula, for the n-th term, is: \[\Large {a_n} = {a_1}{q^{n - 1}}\] so we can write: \[\Large \begin{gathered} {a_6} = {a_1}{q^5} \hfill \\ {a_{12}} = {a_1}{q^{11}} \hfill \\ \end{gathered} \]

  2. Michele_Laino
    • one year ago
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    now, substituting your data into the first formula, we get: \[\Large - 8192 = 8{q^5}\] please solve that equation for q

  3. Michele_Laino
    • one year ago
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    hint: \[\Large q = \sqrt[5]{{\frac{{ - 8192}}{8}}} = ...?\]

  4. anonymous
    • one year ago
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    i got a 4

  5. Michele_Laino
    • one year ago
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    ok! correct!

  6. Michele_Laino
    • one year ago
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    Please wait, I got q=-4 now, substitute q=-4 and a_1=8 into the second equation: \[\Large {a_{12}} = 8 \times {\left( { - 4} \right)^{11}} = ...?\]

  7. anonymous
    • one year ago
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    I got -33554432

  8. Michele_Laino
    • one year ago
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    that's right!

  9. anonymous
    • one year ago
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    Can you help me with this one: Identify the 17th term of a geometric sequence where a1 = 16 and a5 = 150.06. Round the common ratio and 17th term to the nearest hundredth. (2 points) a17 ≈ 123,802.31 a17 ≈ 30,707.05 a17 ≈ 19,684.01 a17 ≈ 216,654.05

  10. Michele_Laino
    • one year ago
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    ok!

  11. Michele_Laino
    • one year ago
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    we can write this: \[\Large {a_5} = {a_1}{q^4}\] so, substituting your data, we have: \[\Large q = \sqrt[4]{{\frac{{{a_5}}}{{{a_1}}}}} = \sqrt[4]{{\frac{{150.06}}{{16}}}} = ...?\]

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