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# In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created

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Math Expert
Joined: 02 Sep 2009
Posts: 44599
In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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28 Feb 2016, 09:17
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In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created by the equation $$a_n=a_{(n−1)}∗2$$. By what percent is the 11th term greater than the 9th term in this sequence?

A. 100%
B. 200%
C. 300%
D. 400%
E. 800%
[Reveal] Spoiler: OA

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Joined: 25 Dec 2012
Posts: 129
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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28 Feb 2016, 11:46
Bunuel wrote:
In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created by the equation $$a_n=a_{(n−1)}∗2$$. By what percent is the 11th term greater than the 9th term in this sequence?

A. 100%
B. 200%
C. 300%
D. 400%
E. 800%

Since its a GP, the property will hold true for the entire sequence. For instance the percent difference between 1st term and 3rd term is equal to the question stem.

Hence 4/1 = 4 which is 400%

IMO D.
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Joined: 12 Aug 2015
Posts: 2580
GRE 1: 323 Q169 V154
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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09 Mar 2016, 04:21
2
KUDOS
Its a Gp series so 11 term = a^10 ; 9th term = a^8
so difference = 3*2^8 => percent gap = 2^8*3/2^8 *100 = 300 %
hence c is correct
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Manager
Joined: 14 Oct 2012
Posts: 175
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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01 Sep 2016, 12:01
1, 2, 4, 8, 16, 32, 64, 128, 256(9th term), 512, 1024(11th term)
Therefore, [(2^10 – 2^8)/2^8]*100 = [2^2 – 1]*100 = 300%
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In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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08 Oct 2017, 06:42
1
KUDOS
Bunuel wrote:
In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created by the equation $$a_n=a_{(n−1)}∗2$$. By what percent is the 11th term greater than the 9th term in this sequence?

A. 100%
B. 200%
C. 300%
D. 400%
E. 800%

$$a_1 = 1, a_2 = 2, a_3 = 4$$

Using $$a_n=a_{(n−1)}∗2$$

$$a_1 = 1$$
$$a_2 = 2$$
$$a_3 = 4 = (2 * 2)$$
$$a_4 = 8 = (2 * 2 * 2)$$
$$a_5 =16 = (2 * 2 * 2 * 2)$$
$$a_6=32=(2 * 2 * 2 * 2*2)$$

By definition, a geometric sequence has a common ratio.

The geometric expansion from $$a_2$$ to $$a_4$$ has the same ratio as the geometric expansion from $$a_{212}$$ to $$a_{214}$$, or the expansion between any two terms, 9$$^{th}$$ and 11$$^{th}$$ included.

In other words, $$a_{n+2}$$ will be the same percent greater than $$a_{n}$$ no matter which terms you choose.

So: $$a_4$$ is what percent greater than $$a_2$$?

Percent increase = $$\frac{New-Old}{Old} * 100$$

$$\frac{8 - 2}{2} =\frac{6}{2} * 100$$ = 300 percent

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Formerly genxer123

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Joined: 11 Jun 2017
Posts: 33
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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08 Oct 2017, 10:53
KEY WORD: Greater than...
Intern
Joined: 07 Aug 2017
Posts: 4
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created [#permalink]

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11 Oct 2017, 14:23
I would just say that every GM sequence within a GM sequence holds the same percentage difference - so difference between first and third term is same as the difference between 9th and 11th term
--> therefore 300%
Re: In the sequence 1, 2, 4, .... an, every nth term for n > 1 is created   [#permalink] 11 Oct 2017, 14:23
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