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 and n>=3, A(1)=1,A(2)=1. Calculate S(1000), where S(1000) is the sum of the first 1000 terms.Options:
-1
0
2
1

• ANS:1
  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(9)=A(8)-A(7)=1-1=0
  A(10)=A(9)-A(8)=0-1=-1
  Ex:S(10) Terms result=> 1+1+0-1-1+0+1+1+0-1=1
  hence S(1000)=1
• 9 years agoHelpfull: Yes(24) No(1)
• 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
  so we can see that after 6 terms we get zero
  A(997)=1
  A(998)=1
  A(999)=0
  A(1000)=-1
  hence S(1000)=1
  ans=1
  
• 9 years agoHelpfull: Yes(15) No(2)
• 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
  
  Here we can see that after every 6 terms series is repeating.
  And the sum of every 6 terms is 0 .. SO,
  S(996)=0;
  S(1000)=S(996)+1+1+0-1
  Which is equal to 1
  Ans= 1
• 9 years agoHelpfull: Yes(4) No(0)
• sol:1
  A(3)=A(2)-A(1)=1-1-0
  similarly,
  A(4)=-1;
  A(5)=-1;
  A(6)=0
  the ans goes as,1 1 0 -1 -1 0 1 1 0...........
  addition of 6 terms is 0.
  s(1000)=996+4=0+(1+1+0+(-1))=1
  
• 9 years agoHelpfull: Yes(3) No(2)
• Answer : 0
  The sequence would be 1 1 0 -1 -1 1 1 0 -1 -1
  so sum would be 0
  
• 9 years agoHelpfull: Yes(2) No(3)
• 1 is answer
  Sequence formed: 1,1,0,-1,-1,0,1,1....repetition after every 6 digits.
  Each 6 digits have sum=0. So 1000mod6=4 digits will remain with effective sum which is:1+1+0+(-1)=1.
• 9 years agoHelpfull: Yes(1) No(1)
• @Khushboo because N>=3 I think s(1000)-a(1) -a(2) =1-2= -1
  Dont you think this is needed?
• 9 years agoHelpfull: Yes(0) No(1)
• 1
  series is
  1,-1,-2,-1,1,2,1,-1,-2,-1,1....
  sum of them will be 0+1(1000th term is 1)
  so s(1000)=1
• 9 years agoHelpfull: Yes(0) No(1)
• ans. =1
  since on writing its terms we get: 1,1,0,-1,-1,0,1,1,0...
  as we can see sum of six consecutive digits is 0 and dividing 1000 by 6 we get 4 as remainder therefore taking four terms say 1,1,0,-1 and taking their sum out we get 1 as the answer.
• 9 years agoHelpfull: Yes(0) No(1)
• series is 1,1,0,-1,-1,0,1,1,0,-1,-1,0....
  so 1,1,0,-1,-1,0 repeats...
  for first 1000 terms (1000/6=166.something)i.e, 166 cycles will fully complete
  so 166*6=996 terms will have sum 0 as sum of each cycle is 0(-1+-1+0+1+1+0=0)
  so last 4 terms will be 1,1,0,-1 whose sum is 1.
• 9 years agoHelpfull: Yes(0) No(0)
• 1
  sum of every 6 terms is 0
• 9 years agoHelpfull: Yes(0) No(0)