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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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Kudos [?]: 128443 [0], given: 12173

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

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19 Sep 2017, 01:17
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45% (medium)

Question Stats:

73% (02:02) correct 27% (02:17) wrong based on 22 sessions

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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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Kudos [?]: 128443 [0], given: 12173

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Joined: 10 Oct 2016
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Kudos [?]: 875 [0], given: 53

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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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19 Sep 2017, 02:02
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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. $$64*\sqrt[3]{3^9}$$

$$A_1=64$$
$$A_2=A_1 * k$$
$$A_3=A_2 * k$$
$$A_4=A_3 * k$$

$$...$$

$$A_n=A_{n-1} * k$$
________________________
$$A_n=64 * k^{n-1}$$

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

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

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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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