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A certain series is defined by the following recursive rule: Sn=K(Sn-1

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A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 06 Jun 2011, 13:02
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A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?

A. ROOT 2
B. ROOT 3
C. 64*ROOT 3
D. 64*3^1/3
E. 64*3^24
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 06 Jun 2011, 14:20
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dimri10 wrote:
:lol: A certain series is defined by the following recursive rule: Sn=K(Sn-1) , where k is a constant. If the 1st term of this series is 64 and the 25th term is 192, wha is the 9th term?
1. ROOT 2
2. ROOT 3
3. 64*ROOT 3
4> 64*3^1/3
5.64*3^24


For GP:
\(A_n=A_1*r^{(n-1)}\)

Here, r=k
\(A_{25}=A_1*k^{(25-1)}=A_1*k^{24}\)

\(192=64*k^{24}\)

\(192=64*(k^{8})^{3}\)

\((k^8)^3=\frac{192}{64}=3\)

Taking cube-root on both sides,
\(k^8=\sqrt[3]{3}\)

Now,
\(A_9=A_1*k^8\)

\(A_9=64\sqrt[3]{3}\)

Ans: "D"
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 06 Jun 2011, 17:52
1
Fluke what does GP stand for? And what is the name of the formula you are referencing?
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 06 Jun 2011, 20:06
Sn = k Sn-1

S25 = 192 = k S24
= k^24 * 64
And S1 = 64 = k S0

==> k^24 = 192/64 = 3.. (1)

Now, S9 = k^8 * 64
From (1), (k^8)^3 = 3
k^8 = 3^\frac{1}{3}
S9 = 3^\frac{1}{3} * 64

Answer D.
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 06 Jun 2011, 23:32
\(S1 = xk = 64
S25 = xk^25 = 192
xkk^24 = 192
k^24 = 192/64 = 3
k = 3^(1/24)
S9 = xk^9=xkk^8 = 64*(3^(1/24*8) = 64*(3^(1/3))\)
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 07 Jun 2011, 00:01
olivite wrote:
Fluke what does GP stand for? And what is the name of the formula you are referencing?


GP is Geometric Progression, where the ratio between two consecutive terms is always same.

According to the question, this series is a Geometric Series.

As,
\(A_2=A_1*k \hspace{3} OR \hspace{3} \frac{A_2}{A_1}=k\)
\(A_3=A_2*k \hspace{3} OR \hspace{3} \frac{A_3}{A_2}=k\)
\(A_4=A_3*k \hspace{3} OR \hspace{3} \frac{A_4}{A_3}=k\)
\(A_5=A_4*k \hspace{3} OR \hspace{3} \frac{A_5}{A_4}=k\)

For such series, the \(n^{th}\) term can be found using following formula:

\(A_n=A_1*k^{(n-1)}\)

Where,
\(k=\)ratio between two consecutive terms
\(A_1=\)first term of the series
\(n=\)index of the term we are trying to find.

Thus,
\(25^{th}\) term of the series, \(A_{25}=A_1*k^{(25-1)}\)
\(9^{th}\) term of the series, \(A_{9}=A_1*k^{(9-1)}\)
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 11 Jun 2011, 09:55
Ans...D

No need for any GP formula here
The rule is that nth term is K times the (n-1)th term.

1st = 64
2nd = k.64
3rd = k^2.64
.
.
.
9th term = k^8 *64
.
.
.
so 25th = k^24*64

Using this solve for k and substitute k in the equation for the 9th term
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 11 Jun 2011, 10:05
s(n) = k s(n-1)

=> its GP = a + ar + ar^2.....+ar^(n-1)


first term = 64 = a
nth term in a GP is given by ar^(n-1)

25th term is 192 = 64* r^24 => r^24 = 3 => r = 3^(1/24)

9th term = ar^8 = 64* 3^(8/24) = 64*3^(1/3)

Answer is D.
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1  [#permalink]

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New post 14 Nov 2018, 23:19
S1 = 64. We have to find out S9 which is equal to k^(8)*64

S25 = k^(24)*S1
Therefore, 192 = k^(24)*64
k = 3^(1/24)

S9 = 3^(8/24)*64 which is 3^(1/3)*64. Option D
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Re: A certain series is defined by the following recursive rule: Sn=K(Sn-1 &nbs [#permalink] 14 Nov 2018, 23:19
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