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chetan2u:
thanks. :)
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Bunuel
In the sequence 2001, 2002, 2003, ..., each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is 2001 + 2002 - 2003 = 2000. What is the 2004th term in this sequence?

(A) -2004
(B) -2
(C) 0
(D) 4003
(E) 6007

series has
4th term= 2000
5th term = 2005
6th term= 1998
7th = 2007

defining a pattern here is a task chetan2u ; please advise on how to solve this question of such a complex pattern

Hi,

Half the work has been done by you..
Now the series is 2001,2002,2003,2000,2005,1998,2007...
So two sequences go on simultaneously..
1) the odd number sequence..
the 1st number is 2001, 3rd is 2003, 5th is 2005
So pattern is the number +2000, thus 27th number will be 2000+27=2027, the 101st will be 2000+101=2101 and so on
2) the even number sequence...
The 2nd number is 2002, 4th is 2000, 6th is 1998..
So the pattern is the number in the list +the value is 2004..
2nd is 2004-2=2002,...4th is 2004-4=2000,...6th is 2004-6=1998

Ok we are looking for 2004th number. This is even and will follow even sequence. Hence 2004-(2004)=0

C
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The sequence is 2001, (2002), 2003, (2000), 2005, (1998), 2007,.........

Now notice the even terms - They are in AP such that d = -2
Also 2004th term in the entire sequence would be 2004/2 = 1002th term in the sequence of even terms

an = a+ (n-1)d
= 2002 + (1002-1)(-2)
= 2002 -2002 = 0
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