Latest Elitmus Aptitude Question SOLUTION: 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.

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(#M40038853) ELITMUS Category question Sequence & Series Keep an EYE

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.

Asked In Elitmus Ravi Ranjan (9 years ago)

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DIMPLE SINGH  (9 years ago)

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

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Ram  (9 years ago)

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

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Jyoti Kumari  (9 years ago)

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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Ayesha Shoaib  (9 years ago)

Is there any min number of quest to b answered??

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