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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
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the general formula, for the nth term, is: \[\Large {a_n} = {a_1}{q^{n  1}}\] so we can write: \[\Large \begin{gathered} {a_6} = {a_1}{q^5} \hfill \\ {a_{12}} = {a_1}{q^{11}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, substituting your data into the first formula, we get: \[\Large  8192 = 8{q^5}\] please solve that equation for q

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1hint: \[\Large q = \sqrt[5]{{\frac{{  8192}}{8}}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Please wait, I got q=4 now, substitute q=4 and a_1=8 into the second equation: \[\Large {a_{12}} = 8 \times {\left( {  4} \right)^{11}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we 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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