if a base 1=139/3, a base 2= -19 and a base k= a base k-1 - a base k-2, then what will be the 62th term in the series? (k, k-1, k-2 are in base)

139/3
-19
-196/3
-19-139/3(-19-196/3)

• a 1=139/3, a 2=-19, a 3=a 2-a 1=-19-139/3=-196/3, a 4=a 3-a 2=-196/3-(-19)=-139/3, a 5=a 4-a 3=-139/3-(-196/3)=57/3, a 6=a 5-a 4=57/3-(-139/3)=196/3, a 7=a 6-a 5=196/3-57/3=139/3=a 1, a 8=a 7-a 6=139/3-196/3=-57/3=-19=a 2 The series repeats after every 6 Numbers. Thus 62th term=6*10+2=2nd number= -19
• 7 years agoHelpfull: Yes(13) No(3)
• Let B as base.
  +aB62
  aB61-aB60
  aB60-aB59-aB60
  -aB59
  -(aB58-aB57)
  -(aB57-aB56-aB57)
  +aB56
  and so on..
  +62 -59 +56 -53.....
  For even value +ve
  For odd value -ve
  Then
  Tn= 62+(n-1)(-3)
  Tn=59-3n
  For 3n
• 6 years agoHelpfull: Yes(0) No(2)