if a1=139/3, a2= -19 and ak= ak-1 - ak-2 , then what will be 62th term in the series ? (k, k-1, k-2 are in base of a)
a. 139/3
b. -19
c. -196/3
d. -19-139/3(-19-196/3)

• a1=139/3
  a2=-19
  a3=a2-a1=-19-139/3=-196/3
  a4=a3-a2=-196/3-(-19)=-139/3
  a5=a4-a3=-139/3-(-196/3)=57/3
  a6=a5-a4=57/3-(-139/3)=196/3
  a7=a6-a5=196/3-57/3=139/3=a1
  a8=a7-a6=139/3-196/3=-57/3=-19=a2
  The series repeats after every 6 numers.
  Thus 62nd number=6*10+2=2nd number= -19
• 7 years agoHelpfull: Yes(32) No(0)
• Given:-
  a1=139/3
  a2=-19
  ak= ak-1 - ak-2
  Now Put (K=3,4,5,6,7,8.......);
  a3=a2-a1=-19-139/3=-196/3
  a4=a3-a2=-196/3-(-19)=-139/3
  a5=a4-a3=-139/3-(-196/3)=57/3
  a6=a5-a4=57/3-(-139/3)=196/3
  a7=a6-a5=196/3-57/3=139/3=a1
  a8=a7-a6=139/3-196/3=-57/3=-19=a2
  Now we can easily see cycle is repeatng after 6 terms
  So 62%6=2 (Remainder)
  Here remainder is 2
  and second term of cycle is -19
  Ans:-> -19
• 7 years agoHelpfull: Yes(4) No(0)
• a1=139/3
  a2=-19
  a3=-19-139/3
  a4=-139/3
  a5=19
  a6=19+139/3
  a7=139/3
  a8=-19
  a9=-19-139/3
  ......
  a2=a8=a14=...=a62=-19
• 7 years agoHelpfull: Yes(1) No(1)
• a1=139/3
  a2=-19
  a3=-19-139/3
  a4=-139/3
  a5=19
  a6=19+139/3
  a7=139/3
  a8=-19
  
  
  Here cyclicity=6;
  
  so a62=a2=-19
• 7 years agoHelpfull: Yes(1) No(1)