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 an integer and n>=3, A(1)=1,A(2)=1. Calculate S(1000), where S(1000) is the sum of first 1000 terms
(1)2
(2)3
(3)4
(4)0

• A(1)=1
  A(2)=1
  A(3)=0
  A(4)=-1
  A(5)=-1
  A(6)=0
  A(7)=1
  hence the series repeats for every 6 terms..
  now divide 1000 by 6=> remainder is 4
  hence S(1000)=A(1)+a(2)+a(3)+a(4)=1+1+0-1=1
  
• 12 years agoHelpfull: Yes(13) No(0)
• A(1)=1 ,A(2)=2
  A(3)=A(2)-A(1)=1
  A(4)=A(3)-A(2)=-1
  A(5)=A(4)-A(3)=-2
  A(6)=A(5)-A(4)=-1
  after 6th term it repeats itself from 1 to 6
  therefore 1000/6=166*6+4
  so A(1)+A(2)+A(3)+A(4)
  S(1000)=1+2+1+-1
  =3
• 12 years agoHelpfull: Yes(3) No(26)
• piyush - a(2)= 1
  you took it 2
  thats y u are wrong
  a(3)= 0
  a(4)= -1
  a(5)= -1
  a(6)= -2
  a(7)= -3
• 12 years agoHelpfull: Yes(2) No(4)
• repeats series after 4 values:
  a(1)=1
  a(2)=2
  a(3)=1
  a(4)=-1
  a(5)=-2
  a(6)=-1
  a(7)=1
  a(8)=0
  a(9)=-1
  a(10)=-1
  a(1)+....+a(10)=-1
• 11 years agoHelpfull: Yes(0) No(3)