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

• 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
• 10 years agoHelpfull: Yes(19) No(0)
• 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
• 10 years agoHelpfull: Yes(2) No(0)
• 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
  
• 10 years agoHelpfull: Yes(1) No(0)