if n>0
a(n)=1/log base n ^2002 then what is the value of a1+a2+a3+a4-a11-a12-a13-a14..?

• Anant look from line-
  =log base 2002^1 + log base 2002^2 + log base 2002^3 + log base 2002^4 + log base 2002^5 - log base 2002^11 -log base 2002^12 -log base 2002^13 -log base 2002^14
  =log base 2002^(1*2*3*4*5)- log base ^(10*11*12*13*14*)
  =log base 2002^(1*2*3*4*5/10*11*12*13*14)
  =log base 2002^(1/11*13*14)
  =log base 2002^(1/2002)
  =-log base 2002^(2002)
  =-1
• 9 years agoHelpfull: Yes(17) No(1)
• question is wrong a1+a2+a3+a4+a5-a10-a11-a12-a13-a14
  answer is -1
  
• 10 years agoHelpfull: Yes(8) No(0)
• a1+a2+a3+a4+a5-a10-a11-a12-a13-a14=?
  using log property log base a^b = 1/ log base b^a
  1/ log base 1^2002 + 1/ log base 2^2002 + 1/ log base 3^2002 + 1/ log base 4^2002 + 1/ log base 5^2002 - 1/ log base 11^2002 - 12/ log base 13^2002 - 1/ log base 14^2002
  = log base 2002^1 + log base 2002^2 + log base 2002^3 + log base 2002^4 + log base 2002^5 - log base 2002^11 -log base 2002^12 -log base 2002^13 -log base 2002^14
  
  u can solve it.....
  n ans is = -1
  
  
• 10 years agoHelpfull: Yes(7) No(0)
• pls solve....how -1
  
• 10 years agoHelpfull: Yes(0) No(0)