Latest Elitmus Aptitude Question SOLUTION: A(n)= 1/(log base n value 2002) what is the value of A(2)+A(3)+A(4)+A(5)-A(10)-A(11)-A(12)-A(13)-A(14) ANS 1 0 2 -1/log2002 3 -1 4 1

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A(n)= 1/(log base n value 2002)
what is the value of A(2)+A(3)+A(4)+A(5)-A(10)-A(11)-A(12)-A(13)-A(14)
ANS
1 0
2 -1/log2002
3 -1
4 1

Asked In Elitmus DEENDAYAL (10 years ago)

Unsolved Read Solution (3)

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RAKESH  (10 years ago)

A(n)= 1/(log base n value 2002)
A(n)= log base 2002 value(n) [base change]

A(2)+A(3)+A(4)+A(5)-A(10)-A(11)-A(12)-A(13)-A(14)
=log base 2002 value(2*3*4*5/10*11*12*13*14)
=log2002(1/11*13*14)
=log2002(1)-log2002(2002)
=0-1
=-1

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Manish kumar sah  (10 years ago)

A(n)=logbase2002value(n)
A(2)+A(3)+A(4)+A(5)-A(10)-A(11)-A(12)-A(13)-A(14)=log2002(2*3*4*5/10*11*12*13*14)
=log2002(1/11*13*14)
=log2002(1/2002)
=-1

Is this solution Helpfull? Yes (2) | No   

sudhir  (10 years ago)

ans is -1
A(n)= 1/(log base n value 2002)
A(2)+A(3)+A(4)+A(5)-A(10)-A(11)-A(12)-A(13)-A(14)
Lets solve as
write A(2)=log2/log2002
bcoz (log base 2 value n=logn/log2)
A(2)=log2/log2002
A(3)=log3/log2002
A(4)=log4/log2002
A(5)=log5/log2002
A(10)=log10/log2002
A(11)=log11/log2002
A(12)=log12/log2002
A(13)=log13/log2002
A(14)=log14/log2002
now write as
A(2)+A(3)+A(4)+A(5)-[A(10)+A(11)+A(12)+A(13)+A(14)]

[ {log2 +log3+log4+log5 ]/log2002}-
{log10+log11+log12+log13+log14}/log2002 ]=-{log11+log13+log14}/log2002
= -1 ans

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