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Math Expert V
Joined: 02 Sep 2009
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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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Difficulty:   55% (hard)

Question Stats: 71% (02:34) correct 29% (02:41) wrong based on 108 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. $$\sqrt{3}$$

B. $$\sqrt{3^9}$$

C. $$64*\sqrt{3^9}$$

D. $$64*\sqrt{3}$$

E. $$64*\sqrt{3^9}$$

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Retired Moderator V
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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. $$\sqrt{3}$$

B. $$\sqrt{3^9}$$

C. $$64*\sqrt{3^9}$$

D. $$64*\sqrt{3}$$

E. $$64*\sqrt{3^9}$$

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

$$...$$

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

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

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

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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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$$A_1$$=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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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. $$\sqrt{3}$$

B. $$\sqrt{3^9}$$

C. $$64*\sqrt{3^9}$$

D. $$64*\sqrt{3}$$

E. $$64*\sqrt{3^9}$$

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

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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   [#permalink] 22 Oct 2018, 07:57
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