consider the sequence of numbers a1,a2,a3....to infinity where a1=81.33 and a2=-19 and aj=(aj-1)-(aj-2).what is the sum of first 6002 terms of this sequence

• x,y,y-x,-x,-y,x-y,x,y....
  x=81.33 y= -19
  after 6 term it will repeat same sequence
  sum of 1st 6 term is 0
  so sum of 6002 term is = sum of (6002 mod 6) term= sum of 2 term= 81.33-19=62.33
• 11 years agoHelpfull: Yes(1) No(0)
• clearly the given sequence is in AP.
  so sum of n terms of an ap is,
  Sn=n/2*{2a+(n-1)d)
  =(6002)/2*{2(81.33)+(6001*(-100.33))}
  =-1806354927.67
• 11 years agoHelpfull: Yes(0) No(3)
• If we check for the first 6 terms we get it as a1=81.33, a2=-19, a3=-100.33, a4=-81.33, a5=19, a6=100.33,After this the terms keep on repeating. We have 6 terms difference so till 6000 the terms cancel each other and we get as zero the extra 2 terms (6001 and 6002)will be 81.33 and -19
  The sum of the two numbers is 62.33
• 3 years agoHelpfull: Yes(0) No(0)