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

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Intern
Joined: 17 Oct 2011
Posts: 10
Location: Taiwan
GMAT 1: 590 Q39 V34
GMAT 2: 680 Q47 V35
The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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02 Mar 2012, 20:12
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77% (01:54) correct 23% (02:32) wrong based on 413 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 65785
Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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02 Mar 2012, 20:29
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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

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

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

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.

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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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
Math Expert
Joined: 02 Sep 2009
Posts: 65785
Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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04 Jun 2013, 04:54
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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GPA: 3.9
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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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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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WE: Sales (Energy and Utilities)
Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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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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Abhimanyu
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Joined: 03 Feb 2020
Posts: 10
Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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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
Math Expert
Joined: 02 Sep 2009
Posts: 65785
Re: The sequence a(1), a(2), ..., a(n), ... is such that an=4a(n-1} - 3  [#permalink]

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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