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# For a Fibonacci sequence, from third term onwards each term

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Manager
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Kudos [?]: 111 [2] , given: 6

For a Fibonacci sequence, from third term onwards each term [#permalink]  08 Nov 2009, 12:56
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Difficulty:

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Question Stats:

41% (04:15) correct 59% (01:59) wrong based on 66 sessions
For a Fibonacci sequence, from third term onwards each term is the sum previous 2 terms. If the difference in squares of seventh and sixth terms of this sequence is 517, what will be the tenth term of this sequence?

A. 147
B. 76
C. 123
D. can't be determined.
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Aug 2012, 00:47, edited 1 time in total.
Edited the question and added the OA.
Math Expert
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Kudos [?]: 47722 [3] , given: 7132

Re: scary algebra...help plz [#permalink]  08 Nov 2009, 14:15
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papillon86 wrote:
For a fibonacci sequence, from third term onwards each term is the sum previous 2 terms. If the difference in squares of seventh and sixth terms of this sequence is 517,
What will be the tenth term of this sequence?

a) 147
b) 76
c) 123
d) can't be determined.

F7^2-F6^2=517

(F7-F6)*(F7+F6)=517

Prime factorization of 517=11*47.
Hence:
F7-F6=11
F7+F6=47

F7=29 and F6=18
F7+F6=F8=29+18=47
F7+F8=F9=29+47=76
F8+F9=F10=47+76=123

P.S. This question is out of the scope of GMAT.
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Kudos [?]: 287 [0], given: 11

Re: For a Fibonacci sequence, from third term onwards each term [#permalink]  12 Dec 2012, 18:42
The sequence from the 6th term looks like this: x... y... x+y... x+2y... 2x+3y...

$$y^2-x^2=517$$
$$(y-x)(y+x)=(47)(11)$$
$$y+x=47$$
$$y-x=11$$
$$y=29&[m]x=18$$[/m]

10th term: $$2x + 3y = 2(18) + 3(29)=123$$

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Impossible is nothing to God.

Re: For a Fibonacci sequence, from third term onwards each term   [#permalink] 12 Dec 2012, 18:42
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