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# Infinite sequence

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Infinite sequence [#permalink]  21 Apr 2010, 05:49
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89% (01:55) correct 11% (01:49) wrong based on 37 sessions
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
C. 2
D. 12/7
E. 6/7
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Re: various questions [#permalink]  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
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}.

Hope it's clear.
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Re: Algebra Sequence [#permalink]  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: Algebra Sequence [#permalink]  17 Feb 2011, 05:49
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
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.

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Re: Algebra Sequence   [#permalink] 17 Feb 2011, 05:49
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