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Intern  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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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
Math Expert V
Joined: 02 Sep 2009
Posts: 58313
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
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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GMAT 1: 710 Q48 V40 GMAT 2: 740 Q49 V42 Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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

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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
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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Joined: 10 Oct 2012
Posts: 591
Re: In the infinite sequence a1, a2,  [#permalink]

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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)

(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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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

We can let a_1 = x, a_2 = 2x, a_3 = 4x, a_4 = 8x and a_5 = 16x. Thus:

16x - 2x = 12

14x = 12

x = 12/14 = 6/7

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Joined: 04 Sep 2017
Posts: 18
Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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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 $$a_5=2^4?$$
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3536
In the infinite sequence a1, a2, a3, ..., an, each term after the firs  [#permalink]

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

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