Question

Let $x^{18}=y^{21}=z^{28},$ then $3,3{\log }_{y}x,3{\log }_{z}y,7{\log }_{x}z$ are in

• A. H.P
• B. A.P
• C. G.P
• D. None of these

Answer 1

The correct option is B A.P

Let $x^{18}=y^{21}=z^{28}=k$

$\therefore 18\log x=21\log y=28\log z=\log k$

$\therefore \log x=\frac{\log k}{18}=\frac{p}{18}$, where

$p=\log k$

$\log y=\frac{\log k}{21}=\frac{p}{21}$

$\log z=\frac{\log k}{28}=\frac{p}{28}$

Now $3,3{\log }_y^x,3{\log }_{z}y,7{\log }_{x}z$

or $3,\frac{3\log x}{\log y},\frac{3\log y}{\log z},\frac{7\log z}{\log x}$

or $3,\frac{3\times 21}{18},\frac{3\times 28}{21},\frac{7\times 18}{28}$

or $3,\frac{7}{2},4,\frac{9}{2}$

which are in A.P. as $\frac{7}{2}-3=4-\frac{7}{2}=\frac{9}{2}-4$