Sequence And Series Question 339
Question: If $ x{{.}^{\ell n( \frac{y}{z} )}}.{y^{_{^{\ell n{{(XZ)}^{2}}}}}}.,{z^{\ell n( \frac{x}{y} )}}={y^{4,\ell n,y}} $ for any x > 1, y >1 and z > 1, then which one of the following is correct?
Options:
A) $ \ell n,y $ is the GM of $ \ell n,x,,\ell n,x,,\ell n,x $ and $ \ell n,z $
B) $ \ell n,y $ is the AM of $ \ell n,x,,\ell n,x,,\ell n,x $ and $ \ell n,z $
C) $ \ell n,y $ is the HM of $ \ell n,x,,\ell n,x,,\ell n,x $ and $ \ell n,z $
D) $ \ell n,y $ is the AM of $ \ell n,Inx,\ell n,z $ and $ \ell n,z $
Show Answer
Answer:
Correct Answer: B
Solution:
[d] $ {x^{ln( \frac{y}{z} )}}.{y^{ln{{(xz)}^{2}}}}.{z^{ln( \frac{x}{y} )}}={y^{4lny}} $
$ \Rightarrow ln[ {z^{ln( \frac{y}{z} )}} ]+ln,[ {y^{ln{{(xz)}^{2}}}} ]+ln[ {z^{ln( \frac{x}{y} )}} ]=ln[ {y^{4lny}} ] $
$ \Rightarrow [ ln( \frac{y}{z} )ln,x ]+[2ln(xz)lny]+[ ln( \frac{x}{y} )lnz ]=4,{{[lny]}^{2}} $
$ \Rightarrow lnx[lny-lnz]+2lny[lnx+lnz] $ $ +,lnz[lnx-lny]=4{{[lny]}^{2}} $
$ \Rightarrow 3lnx+lnz=4lny $
$ \Rightarrow \frac{lnx+lnx=lnx+lnz}{4}=lny $
$ \therefore $ lny is the AM of lnx, lnx, lnx & lnz.
BETA
