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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, 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: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] 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] New post   18 Jun 2012, 01:30
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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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In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post   Updated on: 08 Oct 2022, 22:05
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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, \(a_1 = 12/14 = 6/7\)

Originally posted by KarishmaB on 18 Jun 2012, 02:50.
Last edited by KarishmaB on 08 Oct 2022, 22:05, edited 1 time in total.
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Re: In the infinite sequence a1, a2, [#permalink] 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] New post   02 Sep 2017, 07:08
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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] 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] New post   24 Feb 2018, 21:47
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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
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: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post   06 Mar 2023, 23:09
if someone gets confused they can do it this way by eliminating answers

2, 4 and 24/7 (which is ~3) are all too big for a1 because for e.g. if a1=2 then a5 would be 32 but a2 would 4 and 32-4=28 and not 12

can either guess between 12/7 and 6/7 and quickly calculate that 12/7 =a1 and a5 would be 192/7 and 192/7-24/7 would not give you 12.

Choose answer E
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Re: In the infinite sequence a1, a2, a3, ..., an, each term after the firs [#permalink] New post   06 Mar 2023, 23:19
JeffTargetTestPrep wrote:
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]
06 Mar 2023, 23:19
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