if a(N)=a(N-1)-a(N-2) where n>=3 & integer. given-a(1)=1,a(2)=1.calculate s(1000) dt ts suymn of 21st 1000 integers.
options-
2
3
4
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• a(1)=1,
  a(2)=1,
  a(3)=0,
  a(4)=-1,
  a(5)=-1,
  a(6)=0,
  sum of all these 6 terms is = 0
  and now it follow this cycle continuously upto a(996)
  so sum upto a(996)=0
  so [a(1)+a(2)+.....a(996)]+a(997)+a(998)+a(999)+a(1000)= [ 0 ]+1+1+0-1
  
  so answer is = 1 NO OPTION MATCHED (but its right answer you can check it with
  make a c program for it )
  
  so summation upto
• 12 years agoHelpfull: Yes(43) No(3)
• Answer is 3.
• 12 years agoHelpfull: Yes(6) No(10)
• ans is 3
  see
  a(1)=1
  a(2)=2
  a(3)=1
  a(4)=-1
  a(5)=-2
  a(6)=-1
  then again a(7)=1 a(8)=2 a(9)=1....so on
  so after 6 terms pattern repeats n sum of 6 terms is 0
  hence till 996 sum is 0
  so,a(997) will be 1
  a(998)=2 a(999)=1 a(1000)=-1
  hence sum is 3
  
• 12 years agoHelpfull: Yes(6) No(10)
• C-PROGRAM FOR QUESTION :
  
  
  #include"stdio.h"
  #include"conio.h"
  void main()
  {
  int a[2000],i,sum=0;
  clrscr();
  a[1]=1;a[2]=1;
  for(i=3;i
• 12 years agoHelpfull: Yes(3) No(0)
• @Neha
  in question a(2)=1 not 2
  so plz read question first
  i know there is 9th 10th & 11th question of TCS openseesame are wrong so plz solve it as it given !
  checkout - https://campus.tcs.com/OpenSeesame/
• 12 years agoHelpfull: Yes(2) No(0)
• #include"stdio.h"
  #include"conio.h"
  void main()
  { int a[2000],i,sum=0;
  clrscr();
  a[1]=1;a[2]=1;
  for(i=3;i
• 12 years agoHelpfull: Yes(0) No(0)
• 12 years agoHelpfull: Yes(0) No(1)