Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40032252) ELITMUS Numerical Ability question Elimus Keep an EYE

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

Asked In Elitmus vivek priyadarshi (10 years ago)

Unsolved Read Solution (4)

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Read Solution (4) :

RAKESH  (10 years ago)

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

Is this solution Helpfull? Yes (68) | No (7)  

vikash kumar  (10 years ago)

non of these , ans will be (o-n)log(b^2)ac or(n-m)log(b^2)ac

Is this solution Helpfull? Yes (3) | No (7)  

Vivek Tiwari  (10 years ago)

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

Is this solution Helpfull? Yes | No (9)  

avinash jha  (8 years ago)

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.

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