Mathematics - Sequence & Series Question with Solution | TestHub

Their reciprocals form an A.P. Since $1/{\log }_{ax}x={\log }_x(ax)=1+{\log }_{x}a$, subtracting 1 shows that ${\log }_{x}a$, ${\log }_{x}b$, ${\log }_{x}c$ form an A.P. Thus $2{\log }_{x}b={\log }_{x}a+{\log }_{x}c={\log }_x(ac)$. Hence $b^2=ac$, so $a,b,c$ form a G.P., option B.