if a,b,c given a G.P. (loga+logb+logc)/log6=6.and (b-a) always given a perfect cube no.what the value of a+b+c

• a,b,c are in g.p. so,
  b^2 = ac
  2logb=loga +logc
  
  (loga+logb+logc)/log6=6
  put value
  (2logb+logb)/log6=6
  3logb=6log6
  logb=2log6
  logb=log6^2
  logb=log36
  b=36
  (b-a) is perfect cube so (36-9)=27
  a=9
  b^2=ac
  c=36*36/9
  c=144
  a+b+c=9+36+144=189
• 6 years agoHelpfull: Yes(19) No(3)
• option please?
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• solve (#M40165599) coins questions
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• a, b, and c are in GP.
  So b=ar
  c=ar^2;
  So log a+logb+logc=loga+log(ar)+log(ar^2)
  =3loga+3logr=3log(ar)
  So 3log(ar)/log6=6
  Log(ar)/log6=2
  ar=36
  b=36
  b-a is a perfect cube, let's say 27
  Therefore 36-a=27
  a=9
  r=4
  a+b+c=9+36+9*4*4=189
• 5 years agoHelpfull: Yes(1) No(0)
• (a)189 (b)49 (c)69 (d)100
• 6 years agoHelpfull: Yes(0) No(0)