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If the infinite sequence a1, a2, a3, ..., an, ..., each term

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If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7
[Reveal] Spoiler: OA

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Last edited by Bunuel on 18 Jun 2012, 01:31, edited 2 times in total.
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7

The formula for calculating \(n_{th}\) term would be \(a_n=2^{n-1}*a_1\) . So:
\(a_5=2^4*a_1\);
\(a_2=2*a_1\);

Given: \(a_5-a_2=2^4*a_1-2*a_1=12\) --> \(2^4*a_1-2*a_1=12\) --> \(a_1=\frac{12}{14}=\frac{6}{7}\).

Answer: E.

Hope it's clear.
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 18 Jun 2012, 02:50
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Stiv wrote:
In the infinite sequence \(a_1\), \(a_2\), \(a_3\),...., \(a_n\), each term after the first is equal to twice the previous term. If \(a_5-a_2=12\), what is the value of \(a_1\)?

A. 4
B. 24/7
C. 2
D. 12/7
E. 6/7


First step for sequence questions is writing down the first few terms.
\(a_2 = 2*a_1\)
\(a_3 = 2*a_2 = 2*2*a_1\)
and so on..
\(a_5 - a_2 = 2*2*2*2*a_1 - 2*a_1 = 14 * a_1 = 12\)
So, \(a_1 = 12/14 = 6/7\)

For more on sequences, check out: http://www.veritasprep.com/blog/2012/03 ... sequences/
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 18 Jun 2012, 02:56
a1 = x
a2 = 2x
a3 = 4x
a4 = 16x
an = 2^(n-1)*x

a5-a2 = 2^4*x - 2x = 12
x(2^4 - 2) = 12
x(16 - 2) = 12
x = 12 / 14 = 6 / 7

E

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 18 Jun 2012, 05:30
Took me 2.5 minutes to solve this easy q. I was confused between substituting the answer and trying algebra :P
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Re: In the infinite sequence a1, a2, [#permalink]

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New post 19 Nov 2013, 03:40
amgelcer wrote:
In the infinite sequence a1, a2, a3, ..., an, ..., each term after the first is equal to twice the previous term. If a5 - a2, = 12, what is the value of a1?

(A)
4

(B)
24/7

(C)
2

(D)
12/7

(E)
6/7

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It is a Geometric Progression, with the common ratio as 2.

Thus, as \(t_n = a*r^{n-1}\) , where a is the first term and r is the common ratio.

\(a_5 = a_1*2^4\)and \(a_2 = a_1*2^1\)

Thus,\(a_5-a_2 = a_1*14 = 12 \to a_1 = \frac{6}{7}\)

E.
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term [#permalink]

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New post 09 May 2016, 10:10
Say a1= x

Then a5= 16x and a2= 2x

16x-2x= 12

x= 12/14= 6/7
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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term   [#permalink] 09 May 2016, 10:10
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