Q. log a,log b,log c are in GP and m,n,o are in AP ,then (o-n)log a+(o-m)log b +(n-m)log is equal to ?

a) (o-n)log abc
b) 0
c) 1

• i think qn should be like this
  if a,b,c are in GP and m,n,o are in AP ,then (o-n)log a+(m-o)log b +(n-m)log is equal to ????
  
  
  
  (o-n)loga+(m-o)logb +(n-m)logc
  loga^(o-n)+logb^(m-o)+logc^(n-m)
  =log[a^(o-n)*b^(m-o)*c^(n-m)]
  =log[a^o/a^n * b^m/b^o *c^n/c^m]
  =log[(a/b)^o *(c/a)^n *(b/c)^m]
  =log[(a/b)^o *(c/a)^n *(a/b)^m] [since a,b,c are in GP so a/b = b/c]
  =log[(a/b)^m+o *(c/a)^n ]
  =log[(a/b)^2n *(c/a)^n] [since m,n,o are in AP so m+o=2n]
  =log[(a^2/b^2 *c/a)^n]
  =log[(ac/b^2)^n]
  =log[1^n] [since a,b,c are in GP so b^2=ac]
  =log(1)
  =0
• 10 years agoHelpfull: Yes(68) No(7)
• non of these , ans will be (o-n)log(b^2)ac or(n-m)log(b^2)ac
• 10 years agoHelpfull: Yes(3) No(7)
• u guys are all wrong, it should be like o-n=o-m=n-m=d(assume)
  then >> dloga+dlogb+dlogc>>> d(loga+logb+logc)>>>d(logabc)..
  now d=o-n, hence (o-n)logabc
• 10 years agoHelpfull: Yes(0) No(9)
• The problem is totally right bt the options are wrong.answer will be 2*(n-0)log ABC. m,n,o are in Ap.so let the sp series be 2,4,6.m=2,n=4,0=6. Now
  O-n=2
  O-m=4 &&
  N-m =2.
  So we can write
  O-m=2*(n-m) consider another sp also u will find the same.
  N compute (o-n)=n-m. Now put the values in the problem u will get the answer.
• 9 years agoHelpfull: Yes(0) No(0)