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

• Answer is 1
  A1 = 1, A2 = 1, A3 = A2 - A1 = 0
  A4 = A3- A2 = -1, A5 = A4 - A3 = -1
  A6 = A5 - A4 = 0,
  similarly
  A7 = 1 ; A8 = 1; A9 = 0 .......AND so on ( series repeats after every six terms)
  so SUM will be 1 + 1 + 0+ ( -1) + (-1) + 0 + 1+ 1+ 0 + ......till 1000 terms
  after sixth term series repeats and sum six terms
  1 + 1 + 0+ ( -1) + (-1) + 0 is zero
  sum will be zero till 996 terms ( 996 divisible by 6)
  after 996 series repeats so next four terms will be 1 1 0 -1
  sum of all 1000 terms will be = sum of 996 terms + sum of last four terms
  = 0 + ( 1+1+0+(-1) )
  = 1
• 10 years agoHelpfull: Yes(23) No(0)
• ans should be 0
• 10 years agoHelpfull: Yes(2) No(2)
• using the above formula
  we get the following sequence
  1,1,0,-1,-1,0,1,1,0,-1,-1,0..............
  1+1+0-1-1+0=0
  Hence considering these 6 term as one & its result is zero.
  hence finding the remainder when 1000 is divided by 6
  remainder=4
  so the sum first 4 term is 1+1+0-1=1
  Hence 1 is the Answer
  
• 10 years agoHelpfull: Yes(2) No(0)
• -500 as common difference is zero
• 10 years agoHelpfull: Yes(0) No(1)
• Ans is 3
  Explaination:For determing sum of 1st 1000 no.s using given equestion : A(n)=A(n-1)-A(n-2); n>=3
  A(3)=A(2)-A(1)
  A(1)=1 and A(2)=2 for sum of first 1000 no.s
  Therefore , A(3)=2-1=1
  Similarly, A(4)=1-2=-1
  A(5)=-1-1=-2
  A(6)=-1,A(7)=1,A(8)=2 and so on...
  The series becomes :1,2,1,-1,-2,-1,1,2,1,-1,-2,-1,.... So on
  The sum of first 6 no.s in the series is 0 and so for the sum of 1000 no.s remainder of (1000/6)=4 and so the starting 4 no.s of series is :1,2,1,-1which are the left sum of the series. The final sum thus becomes(1+2+1+(-1))=3
  
• 10 years agoHelpfull: Yes(0) No(6)
• ans) 1.
  a(1)=1
  a(2)=1
  a(3)=0
  a(4)=-1
  a(5)=-1
  a(6)=0
  a(7)=1
  a(8)=1
  a(9)=0
  a(10)=-1
  a(11)=-1
  sum of first 11 terms is 0.
  1000 terms contain =1000/11=90 terms whose sum is 0 and left with initial 10 terms.
  whose sum is 1.
• 9 years agoHelpfull: Yes(0) No(0)