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In the infinite sequence a1, a2, a3, ..., an, each term after the firs

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In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 21 Apr 2010, 05:49
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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
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 21 Apr 2010, 06:37
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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: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 16 Feb 2011, 05:09
This is clearly a geometric progression.

The nth term of a GP is given by a1 r^(n-1) where r is the ratio between two successive terms

a5 = a1 2^(5-1) = a1 2^4 = 16a1
a2 = a1 2^(2-1) = a1 2^1 = 2a1

a5 - a2 = 12
16a1 - 2a1 = 12
14a1 = 12
a1 = 12/14
a1 = 6/7
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 17 Feb 2011, 05:49
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Baten80 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


Let the \(a1= x\)
therefore, \(a2=2x, a3=4x, a4=8x, a5=16x.\)

It is given that \(a5-a2=12\), that means: \(16x-2x=12\); \(14x=12\), therefore \(x=\frac{6}{7}= a1.\)

Answer is E.
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 05 Sep 2014, 11:21
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In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post 17 Jun 2015, 02:32
ITC 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


Just plug in, start with C:

\(a_1\) to \(a_5\): 2 4 8 16 32 >>> does 32-4 equal 12? No! It's way too big. From C move down and repeat.

E finally fits:
6/7 12/7 24/7 48/7 96/7 >>> 96/7 - 12/7 = 84/7 = 12
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In the infinite sequence a1, a2, a3, ..., an, each term after the firs   [#permalink] 17 Jun 2015, 02:32
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