In a sequence of integers , A(n)=A(n-1)-A(n-2), where A(n) is the n'th term in the sequence, n is an integer and n>=3,A(1)=1,A(2)=1
Calculate S(1000), where S(1000)is the sum of the first 1000 terms
a)2
b)3
c)4
d)0

• actually answer is 1...
  consider the soln...
  given A(1)=1,A(2)=1.
  so, A(3)=A(2)-A(1) acc to the given formula A(n)=A(n-1)-A(n-2)
  A(3)=1-1=0.
  similarliy,A(4)=A(3)-A(2)=0-1=-1.
  A(5)=A(4)-A(3)=-1-0=-1.
  A(6)=A(5)-A(4)=-1-(-1)=0.
  after 6th term seq is repeating like...1,1,0,-1,-1,0...after every 6 terms
  so the sum of these 6 terms are 1+1+0+(-1)+(-1)+0=0.
  now we have to find sum of 1000 terms...now consider in splitted pattern,like
  sum of first 600 terms is 0...(as consider 1,1,0,-1,-1,0 as one sequence,now consider 166 seq i.e 166*6=996 terms)
  so the sum of first 996 terms is 0...after that 4 terms come i.e 1,1,0,-1
  so the sum of last 4 terms = 1+1+0+(-1)=1.
  so according to me 1 is answer....
  plzzz give ur feedback
• 11 years agoHelpfull: Yes(264) No(22)
• Ans. 1 (Not in Option)
  
  A(1)=1
  A(2)=1
  A(3)=1-1=0
  A(4)=0-1=-1
  A(5)=-1-0=-1
  A(6)=-1-(-1)=0
  A(7)=0-(-1)=1
  A(8)=1-0=1
  And so on.... we can clearly see that After every n/6 sum of n no become 0
  
  So till S(996) total will be 0
  A(997)=1
  A(998)=1
  A(999)=0
  A(1000)=-1
  So S(1000) = 1
• 11 years agoHelpfull: Yes(23) No(3)
• answer is 0.
  given A(1)=1,A(2)=1.
  so, A(3)=A(2)-A(1) acc to the given formula A(n)=A(n-1)-A(n-2)
  A(3)=1-1=0.
  similarliy,A(4)=A(3)-A(2)=0-1=-1.
  A(5)=A(4)-A(3)=-1-0=-1.
  A(6)=A(5)-A(4)=-1-(-1)=0.
  after 6th term seq is repeating like...1,1,0,-1,-1,0...
  996th term is 6th term(due to repetetion),1000 th term is 4th one.i.e..-1.
  it is in A.P.s(n)=(n/2)(a+l)where l is last term,a is 1st term.
  (1000/2)(1-1)=0.
  k....0 is tje answer.
• 11 years agoHelpfull: Yes(13) No(26)
• actual answer should 1 according to math calculation but according to open seesame answer will be 3......
  how it would be i dnt knw......
• 11 years agoHelpfull: Yes(10) No(4)
• @ divyank and aashly.....we have to find sum of first 1000 terms instead of 1000th term.....read question again carefully....
• 11 years agoHelpfull: Yes(9) No(7)
• if u calculate A(8)=0
  A(16)=0
  ..
  ..
  ..
  A(1000)=A(8*125)=0
  as the series is repeating
• 11 years agoHelpfull: Yes(8) No(30)
• trial and error
  now A(3)=A(2)-A(1)=1-1=0
  A(4)=A(3)-A(2)=0-1=-1
  A(5)=A(4)-A(3)=-1-0=-1
  A(6)=A(5)-A(4)=-1+1=0
  A(7)=A(6)-A(5)=0-1=-1
  A(8)=A(7)-A(6)=-1-0=-1
  .
  .
  .
  A(1000)=0
  so sum is 0
• 11 years agoHelpfull: Yes(7) No(37)
• but 1 is not even in one of the options of openseesame test of tcs.
• 11 years agoHelpfull: Yes(7) No(3)
• finally tell any one answers yar... so many confusions...
• 11 years agoHelpfull: Yes(5) No(6)
• ans is 1.
  sum of first 6 terms is 0
  sum of next 6 terms is 0
  likewise
  
  1000%6 = 4
  last four terms are 1,1,0,-1
  hence ans is=1.
• 10 years agoHelpfull: Yes(3) No(0)
• -1
  A(1000)=A(1999)-A(1998)
  A(1999)=A(1998)-A(1997)
  .
  .
  A(4)=A(3)-A(2)
  A(3)=A(2)-A(1)............sum up all terms
  -------------
  A(1000)=-A(1)
  =-1(ans)
• 11 years agoHelpfull: Yes(2) No(7)
• here,,it is mentioned that no. of terms are 1000 whose sum have to be calculated and n>=3 so terms will start from 3 to 1003 instead of 1->1000
  hence, 0,-1,-1,0,1,1 so sum of 6 terms is 0
  therefore..sum of first 996 terms will be 0 as 996 is divisible by 6
  now last 4 terms are 0,-1,-1,0
  sum = -2
• 10 years agoHelpfull: Yes(2) No(2)
• the answer of this question should be -1
  also the options in the question are wrong
• 9 years agoHelpfull: Yes(0) No(0)
• actually answer is 0...
  consider the soln...
  given A(1)=1,A(2)=1.
  so, A(3)=A(2)-A(1) acc to the given formula A(n)=A(n-1)-A(n-2)
  A(3)=1-1=0.
  similarliy,A(4)=A(3)-A(2)=0-1=-1.
  A(5)=A(4)-A(3)=-1-0=-1.
  A(6)=A(5)-A(4)=-1-(-1)=0.
  A(7)=0-(-1)=1
  A(8)=1-0=1
  after 6th term seq is repeating like...0,-1,-1,0,1,1,0,-1,-1,0,1,1,0,-1...and so on....
  REMEMBER, n>=3 ...so n starts from 3. NOT from 1. so the 1000th term is A(1003)
  so now the sum of the first 6 terms are 0+(-1)+(-1)+0+1+1=0.
  So, sum of 996 terms is also 0. so A(999)=0
  now the next 4 numbers in the series are-- 0,-1.-1.0
  so the sum upto A(1003) i.e. 1000 terms starting from n=3 is (-2)
• 8 years agoHelpfull: Yes(0) No(0)
• We have A(n) = A(n-1)-A(n-2)
  also given are A(1)=A(2)=1
  so, A(3) will come as 0
  A(4)=-1
  A(5)=-1
  and A(6)= 0
  
  Sum of first 6terms, that is from A(1)+A(2)+...+A(6)= 1+1+0-1-1+0=0
  SO, SUM OF FIRST 600 TERMS = 0
  SO SUM OF FIRST 996 TERMS (as 996 is the last number before 1000 that is a multiple of 6 ) = 0
  Now repeating the cycle we get ,
  
  A(997)=1
  A(998)=1
  A(999)=0
  A(1000)=-1
  
  NOW...
  
  A(1000) = A(999) - A(998)
  A(999) = A(998) - A(997)
  A (997) = A(996)-A(995) NOTICE THAT A(998), A(997), A(996),A(995) AND SO ON WILL GET CANCELLED
  .
  .
  .
  A(5) = A(4) - A(3)
  A(4)= A(3)-A(2)
  A(3)=A(2)-A(1)
  ------------------------------------------------
  
  WE SEE AT THE END OF THE SERIES OF SUM WE HAVE ONLY A(999)-A(1) AS THE REST OF THE NUMBERS GET CANCELLED.
  WE KNOW A(999)= 0 AND A(1) = 1
  SO A(999)-A(1) = 0-1 = -1 ANSWER.
  
  PLEASE GIVE FEEDBACK GUYS!!
• 8 years agoHelpfull: Yes(0) No(1)
• for every completion of 6 terms the series is repeating so, the greatest 3 digit number that is divisible by 6 is 996
  i.e., upto 996 the sum =0
  rest of the series is 1,1,0,-1
  so, sum=1+1+0-1=1
  But answer is not found in the option.
• 8 years agoHelpfull: Yes(0) No(0)
• ans:1
  not in option
• 6 years agoHelpfull: Yes(0) No(0)