in a sequence of integers, A(n) = A(n-1)- A(n-2)where A(n) is the nth term in the sequence, n is the integer and n>=3, A(1)=1, A(2)=1 calculate S(1000), where S(1000) is the sum of first 1000 terms

• Ans. 1
  the sequence is 1,1,0,-1,-1,0,1,1,0,-1,-1,0,.......
  so terms are repeated after 6th.hence 1000%6 i.e first 996 will be zero.
  the sum of remaining 4 terms is 1
• 12 years agoHelpfull: Yes(23) No(2)
• A(3)=0
  A(4)=-1
  A(5)=-1
  A(6)=0
  A(7)=1
  continuing we get a series of 1,1,0,-1,-1,0,1....
  the series repeates itself for 6 first digits(1,1,0,-1,-1,0)
  so s(1000)=s(6*136)+s(4)
  =s(6)*136+s(4)
  as the sum of 1st 6 digit is 0 and 1st 4 digit is 1
  so =(0*136)+1
  =1 :)
• 12 years agoHelpfull: Yes(10) No(0)
• 1,1,0,-1,-1,0,1,1,0,-1,-1,0.......
  here we see that the sum of first 5 nos. is=0
  and the series is starting from 1
  thus 1000/5=200
  that is no term is remaining if we take a combination of 5 nos, where each combination=0.
  thus the total value=0
  s(1000)=0.
• 12 years agoHelpfull: Yes(5) No(8)
• sum of first 1000 terms will be
  S(1000) = -1
• 12 years agoHelpfull: Yes(4) No(5)
• but options are 2,3,4,0.
  wich s correct???
• 12 years agoHelpfull: Yes(2) No(0)
• options are
  # 2
  # 3
  # 4
  # 0
  1 and -1 are not in the options
• 12 years agoHelpfull: Yes(1) No(0)
• sum of1000 terms will be=1
• 12 years agoHelpfull: Yes(0) No(0)
• -995....as A(3)=0 A(4)=-1 A(5)=-1 A(6)=-2 &SERIES GOES ON so
  1+1+0-1+(-1)+(996-1)(-1)=-995
• 12 years agoHelpfull: Yes(0) No(3)