a,b,c are in gp. log a + log b + log c / log 6 = 6
b-a is the perfect cube . then what will be a+b+c

• since a,b,c are in geometric sequence so,
  if a=m then,
  b=mn and
  c= mn^2
  therefor,
  m*mn*mn^2= m^3*n^3 = (mn)^3
  now acording to question,
  log a+log b+log c/log 6=6
  so,
  logabc/log6=6
  log abc=6 log6
  abc = 6^6
  
  now
  since, abc=(mn)^3
  put the value of abc ,
  6^6= (mn)^3
  mn= 6*6=36=b
  now acccording to question b-a=Perfect cube
  so a=9=m, as (b-a) =27 that is perfect cube of 3.
  therefor n=4
  therefor c=144
  so a+b+c=9+36+144=189 ans.
• 7 years agoHelpfull: Yes(13) No(3)
• Ans is 189.
  
  Let the three numbers in gp be:
  a/r,a,ar
  then
  3loga/log6=6
  loga=2*log6
  a=36;
  
  now no b - a is a perfect cube i.e
  
  36-36/r = cube;
  nearest perfect cube is 27;
  so 36-36/r=27 => r=4,
  
  so no. are 36/4+36+36*4 = 189
• 6 years agoHelpfull: Yes(0) No(0)