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# The sequence a(1), a(2), ..., a(n), ... is such that

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Intern
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The sequence a(1), a(2), ..., a(n), ... is such that [#permalink]

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02 Mar 2012, 21:12
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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
[Reveal] Spoiler: OA

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Re: If a(3)=x, then a(1)=? [#permalink]

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02 Mar 2012, 21:29
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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, $$a_2=4a_{1}-3$$ --> $$\frac{x+3}{4}=4a_{1}-3$$ --> $$a_1=\frac{x+15}{16}$$.

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: If a(3)=x, then a(1)=? [#permalink]

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02 Mar 2012, 21:34
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 [#permalink]

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04 Jun 2013, 05:54
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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Re: The sequence a(1), a(2), ..., a(n), ... is such that [#permalink]

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05 Jun 2013, 07:01
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 [#permalink]

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22 Sep 2016, 19:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The sequence a(1), a(2), ..., a(n), ... is such that [#permalink]

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08 Sep 2017, 22:00
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   [#permalink] 08 Sep 2017, 22:00
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