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

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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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New post 19 Sep 2017, 01:02
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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. \(64*\sqrt[3]{3^9}\)


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

\(...\)

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

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

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New post 17 Oct 2017, 01:00
\(A_1\)=64

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

\(A_3\)=\(A_2\)*k=64*K*K=64*\(k^2\)

\(A_4\)=\(A_3\)*k=64*\(k^3\)

we see a relationship between \(A_n\) and k which is k is \(K^{n-1}\) for any \(A_n\)

with above information

\(A_{25}\)=64*\(k^{24}\)

Now \(A_{25}\) is given=192

therefore 192=65*\(k^{24}\)=\(\frac{192}{65}\)=\(k^{24}\)

or 3=\(k^{24}\) or k=\(3^{1/24}\)

\(A_9\)=64*\(K^8\)=64*{3^1/24}^8=64*3√3
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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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New post 17 Oct 2017, 03:34
1
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. \(64*\sqrt[3]{3^9}\)


As per the given rule for alpha series, it is a geometric progression.

1st term = 64
25th term = \(192 = 64 * k^{24}\)
\(k^{24} = 3\)

We need to find the 9th term i.e. \(64*k^8\)

\(k^8\) is cube root of \(k^{24}\) so it is cube root of 3. So 9th term is 64*cube root of 3

Answer (D)
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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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Re: An "alpha series" is defined by the rule: A_n = A_{n - 1}*k, where k i &nbs [#permalink] 22 Oct 2018, 06:57
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