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A sequence is given by the rule a(n) = |a_{(n−2)}| – |a_{(n−1)}| for a

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New post 07 Jul 2018, 09:39
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A sequence is given by the rule \(a_n = |a_{(n−2)}| – |a_{(n−1)}|\) for all \(n\geq 3\), where \(a_1 = 0\) and \(a_2 = 3\). What is the value of \(a_{99}\)?


A. -297
B. -3
C. 0
D. 3
E. 297

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A sequence is given by the rule a(n) = |a_{(n−2)}| – |a_{(n−1)}| for a  [#permalink]

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New post 07 Jul 2018, 10:11
Bunuel wrote:
A sequence is given by the rule \(a_n = |a_{(n−2)}| – |a_{(n−1)}|\) for all \(n\geq 3\), where \(a_1 = 0\) and \(a_2 = 3\). What is the value of \(a_{99}\)?


A. -297
B. -3
C. 0
D. 3
E. 297


GMAT does not expect us to calculate too much so there has to be a sequence...

\(a_n = |a_{(n−2)}| – |a_{(n−1)}|\) ........\(a_3 = |a_{(3−2)}| – |a_{(3−1)}|.............a_3=|a_1|-|a_2|..........a_3=0-3=-3\) ...
\(a_4 = |a_2| – |a_3|=|3|-|-3|=0\) ..........
\(a_5 = |a_3| – |a_4|=|-3|-|0|=3\) ..........
so the series is 0,3,-3,0,3,-3,.......
we are looking for 99 and the cyclicity happens every 3 numbers so \(a_{99}=a_3=-3\)

B
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Re: A sequence is given by the rule a(n) = |a_{(n−2)}| – |a_{(n−1)}| for a  [#permalink]

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New post 07 Jul 2018, 10:23
Given :
an=|a(n−2)|–|a(n−1)|an=|a(n−2)|–|a(n−1)| for all n≥3n≥3
where, a1=0 and a2=3
a99 = ?

Solving the equation we get:
a3 = -3
a4 = 0
a5 = 3
a6 = -3

See the repeated pattern yet ?
a1, a2, a3, a4, a5, a6 = 0, 3, -3, 0, 3, -3

Since, a99 is the multiple of 3, so a99 should be the third term in the pattern, in this case its -3.

Hence, B.
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Re: A sequence is given by the rule a(n) = |a_{(n−2)}| – |a_{(n−1)}| for a &nbs [#permalink] 07 Jul 2018, 10:23
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