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The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3

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The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 02 Mar 2012, 20:12
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The sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_n=4a_{n-1}-3\) for all integers n>1. If \(a_3\)=x, then \(a_1=\)?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 02 Mar 2012, 20:29
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The sequence \(a_1\), \(a_2\), …, \(a_n\), … is such that \(a_n=4a_{n-1}-3\) for all integers n>1. If \(a_3\)=x, then \(a_1=\)?
A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16

Since, \(a_n=4a_{n-1}-3\) then \(a_3=4a_{2}-3\) --> \(x=4a_{2}-3\) --> \(a_2=\frac{x+3}{4}\).

Similarly, \(a_2=4a_{1}-3\) --> \(\frac{x+3}{4}=4a_{1}-3\) --> \(a_1=\frac{x+15}{16}\).

Answer: E.

Or substitute the value for \(x\), say \(x=5\), then \(a_3=5=4a_{2}-3\) --> \(a_2=2\) --> \(a_2=2=4a_{1}-3\) --> \(a_1=\frac{5}{4}\). Now, just plug \(x=5\) in the answer choices and see which one yields \(\frac{5}{4}\): only E.

Answer: E.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example if you pick \(x=1\) then you get three "correct" options A, C and E. Generally -1, 0, and 1 are not good choices for plug-in method.

Hope it helps.
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 02 Mar 2012, 20:34
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

My Line of thought:
1. What is given to me a(3)=x, and formula.
2. I need to express a(1) from known.
3. a(n) can be expressed as a(n-1) means it can be expressed by any of the predecessor or successor.
as a(n-1) can be expressed as a(n-2)....hence a(n) can be expressed as a(n-2) and vice versa
4. a(3) = 4a(2)-3 , (n=3)>1
= 4 [4a(1)-3]-3 , (n=2)>1
=16a(1)-12-3
x = 16a(1)-15
(x+15)/16 = a(1) == Ans E
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 04 Jun 2013, 04:54
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 05 Jun 2013, 06:01
1
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16


I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 08 Sep 2017, 21:00
1
a(1)= {a(2)+3}/4
Splitting the denominator a(1)= a(2)/4 +3/4 say eq----i
a(2)={a(3)+3}/4
a(2)={x+3}/4
a(2)/4= {x+3}/16 let say this is eq -----ii

Sub a(2)/4 in eq i

a(1)= {x+3}/16 + 3/4
a(1)= (x+15)/16
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 05 Feb 2020, 18:59
ziko wrote:
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16


I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.
please explain me how x+3/4=4a(1)-3 , i uderstand your explanation just not this part
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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New post 05 Feb 2020, 23:29
sujal998 wrote:
ziko wrote:
The sequence a(1), a(2), …, a(n), … is such that a(n)=4a(n–1) –3 for all integers n>1. If a(3)=x, then a(1)=?

A. 4x–3
B. 16x–15
C. (x+3)/4
D. (x+3)/16
E. (x+15)/16


I love such questions!

Since we know a(3) we can find a(2), if we can find a(2) we can find a(1). a(3)=4a(2)-3 ---> x=4a(2)-3 ---> a(2)=(x+3)/4

a(2)=4a(1)-3 ---> (x+3)/4=4a(1)-3 ---> 4a(1)=(x+15)/4 ---> a(1)=(x+15)/16 The answer is E.


please explain me how x+3/4=4a(1)-3 , i uderstand your explanation just not this part


Substitute of \(a_2=\frac{x+3}{4}\) into \(a_2=4a_1-3\)
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Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3   [#permalink] 05 Feb 2020, 23:29

The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3

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