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# For the sequence a1,a2,a3,…an , an=(an−1)2–an−2 . If a1=1 and a2=0 ,

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Joined: 02 Sep 2009
Posts: 50058
For the sequence a1,a2,a3,…an , an=(an−1)2–an−2 . If a1=1 and a2=0 ,  [#permalink]

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17 Apr 2018, 05:21
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Difficulty:

15% (low)

Question Stats:

81% (01:36) correct 19% (01:25) wrong based on 49 sessions

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For the sequence $$a_1$$, $$a_2$$, $$a_3$$, … $$a_n$$, $$a_n=(a_{n−1})^2–a_{n−2}$$. If $$a_1=1$$ and $$a_2=0$$, what is the sixth term of the sequence?

A. 0
B. 1
C. 2
D. 3
E. 7

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Re: For the sequence a1,a2,a3,…an , an=(an−1)2–an−2 . If a1=1 and a2=0 ,  [#permalink]

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17 Apr 2018, 06:26
Bunuel wrote:
For the sequence $$a_1$$, $$a_2$$, $$a_3$$, … $$a_n$$, $$a_n=(a_{n−1})^2–a_{n−2}$$. If $$a_1=1$$ and $$a_2=0$$, what is the sixth term of the sequence?

A. 0
B. 1
C. 2
D. 3
E. 7

Using the formula above

$$a_3$$ = (0)^2 -1 = -1
$$a_4$$ = (-1)^2 -0 = 1
$$a_5$$ = (1)^2 -(-1) = 2
$$a_6$$ = (2)^2 - (1) = 3

Hence option D = 3 is the answer and the sixth term.
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Re: For the sequence a1,a2,a3,…an , an=(an−1)2–an−2 . If a1=1 and a2=0 , &nbs [#permalink] 17 Apr 2018, 06:26
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