In a sequence of integers A(n)=A(n-1)when A(n) is the nth term in the sequence n is an integer and n>=3,A(1)=1,A(2)=1calculate s(1000) where s(1000) is the sum of first 1000 terms ?

• ans would be 1. ques is incmplte it shld be A(n)=A(n-1)-A(n-2)
  putting n=3,4,5.. we'll get A(3)=A(2)-A(1), A(4)=A(3)-A(2) and so on
  S(1000)=1
• 11 years agoHelpfull: Yes(31) No(9)
• This question form a series on putting values and also question is wrong.
  it should be-- A(n)= A(n-1) - A(n-2)
  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
  a(9)= -1+1= 0
  series of first six r=terms comes out to be-- 0,-1,-1,0,1,1= whose sum is zero.
  Please note that here series formulation of six
  there are 966 terms covered by pair of six, thus making sum zero upto 996
  we are left with 1 term ie 997 term.
  therefore sum= 0
  because there are total 997 terms not 1000.
  
  
• 11 years agoHelpfull: Yes(7) No(7)
• A(n)= A(n-1) n>=3(given)
  so n=3,4,5,.......till 1000th term(1003)
  A(3)=A(2)=1
  A(4)=A(3)=1
  A(5)=A(4)=1
  .
  .
  .
  .
  A(1000)=A(999)=1
  A(1001)=A(1000)=1
  A(1002)=A(1001)=1
  A(1003)=A(1002)=1
  we have find sum of first 1000 terms which means...
  1+1+1+1+1...............till 1000th term
  sum=n/2[(2*a)+ (n-1)d]
  = 1000/2[2*1+(999)*0]
  = 1000Ans.
• 11 years agoHelpfull: Yes(5) No(4)
• answer is 1000.....
• 11 years agoHelpfull: Yes(3) No(14)
• A(n)=A(n-1)
  A(1)=1
  a(2)=1
  .
  .
  .
  A(1000)=1
  s(1000)=1
  ans=1
• 11 years agoHelpfull: Yes(3) No(2)
• A(3)=A(2)-A(1)=0,A(4)=A(3)-A(2)=-1,A(5)=A(4)-A(3)=-1,A(6)=A(5)-A(4)=0....IN THIS WAY..IF IF MAKE GROUP OF SIX NOS WE GET SUM =0;;;SO WHEN 1000 TAKEN...THERE IS 166 GROUP THAT MEANS TILL 996 ANS IS 0...SO 997=1,998=1,999=0,1000=-1 SUM OF 4 DIGITS i.e (997th+998TH+999th+1000th)term =(1+1+0-1=1) answer equal to (1)
• 11 years agoHelpfull: Yes(1) No(1)
• A(1)=1
  A(2)=1
  A(3)=A(3-1)=A(2)=1
  A(4)=1
  
  
  lly A(1000)=1
  
  A(1)+A(2)+.....+A(1000)=1+1+1+......+1=1000
• 11 years agoHelpfull: Yes(0) No(1)
• 1 is the answer !!!
• 11 years agoHelpfull: Yes(0) No(0)