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

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For a Fibonacci sequence, from third term onwards each term [#permalink]

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08 Nov 2009, 13:56
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Question Stats:

43% (03:35) correct 57% (01:59) wrong based on 68 sessions

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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, 01:47, edited 1 time in total.
Edited the question and added the OA.

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08 Nov 2009, 15:15
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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 [?]: 128901 [3], given: 12183

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Re: For a Fibonacci sequence, from third term onwards each term [#permalink]

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12 Dec 2012, 19:42
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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Re: For a Fibonacci sequence, from third term onwards each term [#permalink]

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09 Aug 2015, 07:34
Bunuel wrote:
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.

I think the answer should be 4. Cannot be determined as the even 1 * 517 could also be considered.
In that case T6 and T7 would be 258 and 259, giving a different T10

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Re: For a Fibonacci sequence, from third term onwards each term [#permalink]

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09 Aug 2015, 12:56
Yes, Caligula is right that the answer should be 'cannot be determined', at least as the question is worded. They don't even tell us the numbers need to be integers, nor that they need to be positive. If you take the first solution above (where 18 and 29 are the sixth and seventh terms), and just make those numbers negative, you'll get another sequence that 'works', and -123 can be the tenth term. And if you allow non-integers in the sequence, there are infinitely many possibilities.

The question only has a unique answer if you know the terms must all be positive integers. Then the other factorization (1)(517) doesn't work, because that would force some of the earliest terms in the sequence to be negative. Either way, it's not a realistic GMAT question at all, so test takers shouldn't worry about it.
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Kudos [?]: 1953 [0], given: 6

Re: For a Fibonacci sequence, from third term onwards each term   [#permalink] 09 Aug 2015, 12:56
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