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

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Joined: 21 Apr 2010
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In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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21 Apr 2010, 05:49
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81% (01:37) correct 19% (02:03) wrong based on 195 sessions

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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
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Posts: 51307
Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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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: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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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]

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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
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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In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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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
C. 2
D. 12/7
E. 6/7

$$a_1$$ 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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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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12 Jun 2017, 18:19
16a1-2a1 = 12 gives a straight forward answer as 6/7
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Joined: 04 Sep 2017
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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24 Feb 2018, 15:30
Bunuel 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
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.

Can someone help me out here? I dont know if I am reading the question correctly but it says each term after the first term is equal to twice the previous term but the solution above shows it as two times the previous term times the first term. I am not seeing how that makes sense?

Shouldnt it be $$a_5=2^4?$$
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In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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24 Feb 2018, 20:47
teamryan15 wrote:
Bunuel 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
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.

Can someone help me out here? I dont know if I am reading the question correctly but it says each term after the first term is equal to twice the previous term but the solution above shows it as two times the previous term times the first term. I am not seeing how that makes sense?

Shouldnt it be $$a_5=2^4?$$

teamryan15
I think I see where you're a little off. The first term is not 1. You are focused on just the coefficient / multiplier, I think.

If $$A_1$$ were 1, then yes, $$A_5$$ would = $$2^4$$. We would have:
$$A_1 = 1$$
$$A_2 = 2$$
$$A_3 = 4$$
$$A_4 = 8$$
$$A_5 = 16 = 2^4$$

The first term is $$a$$, not 1. Thus:

$$A_1 = a_1$$
$$A_2 = (2*a_1) = 2a_1 = 2^1*a_1$$
$$A_3 = (2*2a_1)= 4a_1 = 2^2*a_1$$
$$A_4 = (2*4a_1)= 8a_1 = 2^3* a_1$$
$$A_5 = (2*8a_1) = 16a_1 = 2^4*a_1$$

Hope that helps.
In the infinite sequence a1, a2, a3, ..., an, each term after the firs &nbs [#permalink] 24 Feb 2018, 20:47
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