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Updated on: 05 Aug 2012, 01:47
Originally posted by papillon86 on 08 Nov 2009, 13:56.
Last edited by Bunuel on 05 Aug 2012, 01:47, edited 1 time in total.
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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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.
A. 147
B. 76
C. 123
D. can't be determined.
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08 Nov 2009, 15:15
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papillon86
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
Answer: C.
P.S. This question is out of the scope of GMAT.
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12 Dec 2012, 19:42
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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: \(2x + 3y = 2(18) + 3(29)=123\)
Answer: 123
\(y^2-x^2=517\)
\((y-x)(y+x)=(47)(11)\)
\(y+x=47\)
\(y-x=11\)
\(y=29&[m]x=18\)[/m]
10th term: \(2x + 3y = 2(18) + 3(29)=123\)
Answer: 123
