In a sequence of integer,A(n)=A(n-1)-A(n-2),where n is the nth term of 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

• given A(n)=A(n-1)-A(n-2) & A(1)=1 A(2)=1
  now A(3)=A(3-1)+A(3-2)=1-1=0
  similarly, A(4)=0-1=-1, A(5)=-1-0=-1, A(6)=-1+1=0..
  from the 7th term the same sequence repeats.
  so, first we cacl sum of first 6 tems 1+1+0-1-1+0=0
  now we have to find sum of 1000 terms, for this, we divide 1000/6, there will be 4 terms left, the remaining 996 terms sum is 0. those 4 terms are 1,1,0,-1.
  now sum of these 4 terms is 1. so, the ans is 1.
• 12 years agoHelpfull: Yes(51) No(3)
• answer is 3
• 12 years agoHelpfull: Yes(3) No(12)
• Ans=1
  sum of first 10 terms = 1
  sum of second 10 terms = 1
  sum of third 10 terms = -2
  1000/30=33 + 10 as remainder
  sum=0+1=1
• 12 years agoHelpfull: Yes(3) No(0)
• ans is 1,series will b 1 1 0 -1 -1 0 1 1 0 -1 -1 0 0 1 1....so sum of first six term is 0 till 996 terms sum will be zero,remaining will be 1 1 0 -1 sum is 1
• 12 years agoHelpfull: Yes(3) No(0)
• A1=1
  A2=1
  A3=0
  A4=-1
  A5=-1
  A6=0
  A7=1
  .
  .
  .
  REM(1000/6)=4
  i.e. SUM(1000)=SUM(4terms)
  sum(1000)=1+1+0-1
  =1(Ans)
• 12 years agoHelpfull: Yes(2) No(1)
• a(n)=a(n-1)+a(n-2)
  so, a(3)=a(2)-a(1)
  a(4)=a(3)-a(2)
  a(1000)=a(999)-a(1)
  also s(4)=a(3)-a(1)
  s(5)=a(4)-a(1)
  so
  s(1000)=a(999)-a(1) as remaining terms gets cancelled
  now a(999)=0 as 999/6 will give reminder as 3. then s(1000)=0-1=-1.
  
• 12 years agoHelpfull: Yes(0) No(4)
• s(1000)=s(999)-s(998)
  =s(998)-s(997)-s(998)
  =-s(997)=-(s(996)-s(995))
  finaly we got s(1) which will become to 1
• 12 years agoHelpfull: Yes(0) No(0)