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An "alpha series" is defined by the rule: A_n = A_{n - 1}*k, where k i

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An "alpha series" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]

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An "alpha series" is defined by the rule: \(A_n = A_{n - 1}*k\), where k is a constant. If the 1st term of a particular "alpha series" is 64 and the 25th term is 192, what is the 9th term?


A. \(\sqrt[24]{3}\)

B. \(\sqrt[24]{3^9}\)

C. \(64*\sqrt[24]{3^9}\)

D. \(64*\sqrt[3]{3}\)

E. \(64*\sqrt[3]{3^9}\)
[Reveal] Spoiler: OA

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Re: An "alpha series" is defined by the rule: A_n = A_{n - 1}*k, where k i [#permalink]

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Bunuel wrote:
An "alpha series" is defined by the rule: \(A_n = A_{n - 1}*k\), where k is a constant. If the 1st term of a particular "alpha series" is 64 and the 25th term is 192, what is the 9th term?


A. \(\sqrt[24]{3}\)

B. \(\sqrt[24]{3^9}\)

C. \(64*\sqrt[24]{3^9}\)

D. \(64*\sqrt[3]{3}\)

E. \(64*\sqrt[3]{3^9}\)


\(A_1=64\)
\(A_2=A_1 * k\)
\(A_3=A_2 * k\)
\(A_4=A_3 * k\)

\(...\)

\(A_n=A_{n-1} * k\)
________________________
\(A_n=64 * k^{n-1}\)

We have \(A_{25}=64 * k^{24}=192 \implies k^{24} = 3 \implies k = (3)^{\frac{1}{24}}\)

Hence \(A_9=64 * k^8 = 64 * \Big ((3)^{\frac{1}{24}} \Big ) ^ 8 = 64 * (3)^{\frac{8}{24}} = 64 * (3)^{\frac{1}{3}} = 64 * \sqrt[3]{3}\)

Answer D
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Kudos [?]: 875 [0], given: 53

Re: An "alpha series" is defined by the rule: A_n = A_{n - 1}*k, where k i   [#permalink] 19 Sep 2017, 02:02
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