J1,J2,J3......Jn be a series of n Numbers .And J1 = 139/3 or J2 = (-19) & Jk = J(k-1) - J(k-2). Then Calculate the 602th Term of the series.

• J1= 139/3
  J2= -19
  J3= J2-J1 = -196
  J4= J3-J2 = -139
  J5= J4-J3 = 19
  J6= J5-J4 = 196/3
  
  therefore the series is reapeting as
  139/3 , -19, -196/3, -139/3, 19 , 196/3 ,139/3.......
  600th term will be same as the 6th term ..
  600th term = 196/3
  601th term = 139/3
  602th term = -19
  
• 9 years agoHelpfull: Yes(9) No(0)
• Let k= J3 = J2-J1
  J4 = J3-J2 = (J2-J1)-J2= -J1
  J5 = J4-J3 = (J3-J2)-J3= -J2
  J6 = J5-J4 = (J4-J3)-J4 =-J3 =-k
  J7 = J6-J5 = (J5-J4)-J5 =-J4 =J1
  J8 = J7-J6 = (J6-J5)-J6 =-J5 =J2
  J9 = J8-J7 = (J7-J6)-J7 =-J6 =+k
  
  J(3,6,9,12.........600) = + or - k [3*200=600]
  J(4,6,10.............) = + or - J(1)
  J(5,8,11,14,17,20......602)= + or - J(2)[600 is +-k,601 is +-J(1),602 is +-J(2)]
  and
  J(5,11,17.....601 odd numbers) =-J(2)
  J(8,14,20.....602 even numbers) =+J(2)
  answer is J(602)= J(2)= -19
• 9 years agoHelpfull: Yes(1) No(0)
• ans:- -19
  j(1)=139/3
  j(2)=-19
  j(3)=-196/3
  j(4)=-139/3
  j(5)=19
  602/5 rem=2
  so ans will be j(2)=-19
  
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• Is there any min number of quest to b answered??
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