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Difficulty: 505-555 Level,   Sequences,                           
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Re: Sequence Problem [#permalink]
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Good one. Please let me know if some one comes up with a solution that can be worked under 2 mins :)

given an= 2an-1 - x

a5 = 99 = 2 a4 - X = 2[2a3-X] -X = 4 a3 - 3X
given a3 = 27 ; substituting:

108 - 3X = 99 => X = 3

A
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The sequence a1…..a2.......an........is such that an=2an-1-X for all positive integers n>=2 and for certain number X. If a5=99 and a3=27, what is the value of X?
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Re: Sequence [#permalink]
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monirjewel wrote:
The sequence a1…..a2.......an........is such that an=2an-1-X for all positive integers n>=2 and for certain number X. If a5=99 and a3=27, what is the value of X?


Plug the known values a5=99 and a3 = 27 into the formula:

a5 = 2(a4) - x
99 = 2(a4) - x

a4 = 2(a3) - x
a4 = 2(27)-x = 54-x

Substitute 54-x for a4 in the top equation:
99 = 2(54-x)-x
99=108-3x
3x=9
x=3

On the GMAT, I would recommend that you plug in the answer choices, one of which would say that x=3.

Plug a5 = 99 and x=3 into the formula:
99 = 2(a4) -3
a4 = 51

Plug a4=51, a3=27, and x=3 into the formula:
51 = 2(27) - 3
51 = 51. Success!
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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink]
whiplash2411 wrote:
udaymathapati wrote:
The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?
A. 3
B. 9
C. 18
D. 36
E.45


Quite simple to solve this one.

Given: \(a_n = 2a_{n-1} - x\)

\(a_5 = 99\)

\(a_3 = 27\)

\(a_5 = 2a_4 - x = 2(2a_3 - x) - x = 4a_3 - 3x = 99\)

\(4(27) - 3x = 99\)\(3x = 108-99 = 9\)

\(x = 3\)

QUESTION : How exactly did you get from 2(2a3 - x) to 4a3 * 3x, wouldn't it be 4a3 - 2x?
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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink]
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sagnik242 wrote:
whiplash2411 wrote:
udaymathapati wrote:
The sequence a1, a2, … , a n, … is such that an = 2an-1 - x for all positive integers n ≥ 2 and for certain number x. If a5 = 99 and a3 = 27, what is the value of x?
A. 3
B. 9
C. 18
D. 36
E.45


Quite simple to solve this one.

Given: \(a_n = 2a_{n-1} - x\)

\(a_5 = 99\)

\(a_3 = 27\)

\(a_5 = 2a_4 - x = 2(2a_3 - x) - x = 4a_3 - 3x = 99\)

\(4(27) - 3x = 99\)\(3x = 108-99 = 9\)

\(x = 3\)

QUESTION : How exactly did you get from 2(2a3 - x) to 4a3 * 3x, wouldn't it be 4a3 - 2x?


\(a_4=2a_3-x\) --> \(a_5 = 2a_4 - x\) --> \(a_4=2(2a_3-x)-x=4a_3-2x-x=4a_3-3x\).
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Re: The sequence a1, a2, … , a n, … is such that an = 2an-1 - x [#permalink]
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jet1445 wrote:
The sequence \(a_1\), \(a_2\), … , \(a_n\), … is such that \(a_n = 2a_{n-1} - x\) for all positive integers n ≥ 2 and for certain number x. If \(a_5 = 99\) and \(a_3 = 27\), what is the value of x?

A. 3
B. 9
C. 18
D. 36
E. 45


We can create the equation:

a(5) =2 *a(4) - x

99 = 2 * a(4) - x

(99 + x)/2 = a(4)

and

a(4) = 2 * a(3) - x

a(4) = 2 * 27 - x

a(4) = 54 - x

Substituting (54 - x) for a(4) into the first equation, we have::

(99 + x)/2 = 54 - x

99 + x = 108 - 2x

3x = 9

x = 3

Answer: A
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Re: The sequence a1, a2, , a n, is such that an = 2an-1 - x [#permalink]
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Re: The sequence a1, a2, , a n, is such that an = 2an-1 - x [#permalink]
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