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 then calculate s(1000) where s(1000) is the sum of first 1000 terms

• a(n) = a(n-1) - a(n-2)
  
  a(3) = a(2) - a(1)
  a(4) = a(3) - a(2)
  a(5) = a(4) - a(3)
  a(6) = a(5) - a(4)
  .
  .
  .
  .
  .
  a(1000) = a(999) - a(998)
  
  Add all these equations and look carefully on adding terms are cancelling each other, so at last we will get the final expression as :-
  
  a(1000) = -a(1)
  a(1) + a(1000) = 0
  Adding a(2) on both sides
  
  a(1) + a(2) + a(1000) = a(2)
  
  We have reached this expression by adding all the terms so the expression on left side is nothing but the value of s(1000)
  
  so answer is s(1000) = a(2) = 1
  
  
• 11 years agoHelpfull: Yes(23) No(4)
• we know that a(1)=1, a(2)=1
  from these values we can get a(3),a(4).... and soon.
  but at a(7) the cycle will repeat.
  a(3)=0,a(4)=-1,a(5)=-1,a(6)=0
  then 1000/6=166 with remainder 4. by adding the first four values we can get 1
• 11 years agoHelpfull: Yes(11) No(0)
• we know that
  a(1)=1,
  a(2)=1
  a(3)=0,
  a(4)=-1
  a(5)=-1
  a(6)=0
  but at a(7) the cycle will repeat.
  Hence it has cyclicity of 6.
  Now when we add,
  a(1)+a(2)
  
  a(3) = a(2) - a(1)
  a(4) = a(3) - a(2)
  a(5) = a(4) - a(3)
  a(6) = a(5) - a(4)
  .
  .
  .
  .
  .
  a(1000) = a(999) - a(998)
  All the terms gets cancelled excluding a(2) and a(999)
  Hence s(1000)=a(2)+a(999)
  using cycility of 6 we can calculate a(999)
  ie. divide by 6 get the remainder
  a(999)=0
  Therefore,
  s(1000)=a(2)+a(999)
  = 1+0
  s(1000)=1 ans.
  
• 11 years agoHelpfull: Yes(3) No(2)
• ans is 1....
  
• 11 years agoHelpfull: Yes(2) No(0)
• how can u say a(999)=0 ?
• 10 years agoHelpfull: Yes(2) No(0)
• I want ask to vijeta that the questions are on m4 site are comes on tcs or not .if comes then please send me question papers on id naendra.madhur@gmail.com.
• 11 years agoHelpfull: Yes(1) No(0)
• @vinay jain. on adding the equations we will get a(3)+a(4)+.....a(1000)= -a(1)+a(999)
  on adding a(1)&a(2) on both sides we get
  a(1)+a)2)+a(3)+......a(1000)= a(1)+a(2)-a(1)+a(999)
  s(1000)= a(2)+a(999).
  if i am right please explain me further
• 11 years agoHelpfull: Yes(1) No(0)