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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]

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New post 21 Apr 2010, 06:49
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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: If the infinite sequence a1, a2, a3, ..., an, ..., each term  [#permalink]

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New post 18 Jun 2012, 01:30
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]

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New post 17 Feb 2011, 06:49
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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: 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: In the infinite sequence a1, a2,  [#permalink]

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New post 19 Nov 2013, 03:40
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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)
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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New post 02 Sep 2017, 07:08
alimad 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


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

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

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New post 24 Feb 2018, 16:30
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
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.


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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New post 24 Feb 2018, 21:47
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
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.


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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Re: If the infinite sequence a1, a2, a3, ..., an, ..., each term   [#permalink] 02 Sep 2018, 15:34
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