In a given 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

• Its answer must be 1
  bcoz series is like 1,1,0,-1,-1,0,1,1,0,-1,-1,0,.....
  sum of 996 numbers are 0
  remaining series after 996 numbers are 1,1,0,-1
  its sum is 1
• 11 years agoHelpfull: Yes(39) No(5)
• its answer must be zero
  bcoz series is like 1,1,0,-1,-1,....
  considering this sum of 1000 number will be zero
• 11 years agoHelpfull: Yes(6) No(5)
• the series will like as 1,1,0,-1,-1,0,1,1.....the series is repeating with 6 elements.the sum of every 6 element will be 0.and there are 166 group of 6 element their sum will be zero & remaining 4 element will produce the equals 1
• 11 years agoHelpfull: Yes(4) No(0)
• ans : 0;
  the pattern of sequence is 0,-1,-1,0,1,1 (with sequence starting from n>=3) which repeats after every 6 terms,and sum of each set=0.
  so in the sequence of 1000 terms there are 166 sets of 6 elements,so total sum of first 996 elements is 0 (covering 996 elements).
  last 4 elements of first 1000 will be 0,-1,-1,0 with sum = 0.
  thus ans. is : 0.
• 11 years agoHelpfull: Yes(2) No(3)
• ans is 3
  a(3)=0
  a(4)=-1
  a(5)=1
  .
  .
  .
  .
  .a(10)=1
  hence s(1000)=a(1)+a(2)+a(1000)=1+1+1=3
• 11 years agoHelpfull: Yes(1) No(6)
• ans :-2;
  the pattern of sequence is 0,-1,-1,0,1,1 (important: as sequence starting from n>=3 so A(1) & A(2) not in sequence). Pattern repeats after every 6 terms,and sum of each set=0.
  so in the sequence of 1000 terms there are 166 sets of 6 elements,so total sum of first 996 elements is 0 (covering 996 elements).
  last 4 elements of first 1000 will be 0,-1,-1,0 with sum = -2.
  thus ans. is :-2.
  [mistake in que. is of -ve sign before 2]
• 11 years agoHelpfull: Yes(1) No(3)
• 1+1+0+(-1)=1
• 11 years agoHelpfull: Yes(0) No(1)