Question

If a, b, c are in G.P. and a^1/x = b^1/y = c^1/z, then xyz are in
(a) AP
(b) GP
(c) HP
(d) none of these

Answer 1

(a) AP

$$\begin{gathered} a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore b^2=ac \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides}: \\ 2\log \mathrm{b}=\log \mathrm{a}+\log \mathrm{c}........\left(\mathrm{i}\right) \\ \mathrm{Now},a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{the}\mathrm{side}s: \\ \frac{\log \mathrm{a}}{\mathrm{x}}=\frac{\log \mathrm{b}}{\mathrm{y}}=\frac{\log \mathrm{c}}{\mathrm{z}}........\left(\mathrm{ii}\right) \\ \mathrm{Now},\mathrm{comparing}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right): \\ \frac{\log a}{x}=\frac{\log \mathrm{a}+\log \mathrm{c}}{2y}=\frac{\log \mathrm{c}}{\mathrm{z}} \\ \Rightarrow \frac{\log a}{x}=\frac{\log \mathrm{a}+\log \mathrm{c}}{2y}\mathrm{and}\frac{\log \mathrm{a}}{\mathrm{x}}=\frac{\log \mathrm{c}}{\mathrm{z}} \\ \Rightarrow \log \mathrm{a}\left(2\mathrm{y}-\mathrm{x}\right)=\mathrm{xlog}\mathrm{c}\mathrm{and}\frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}} \\ \Rightarrow \frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{\left(2\mathrm{y}-\mathrm{x}\right)}\mathrm{and}\frac{\log \mathrm{a}}{\log \mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}} \\ \Rightarrow \frac{\mathrm{x}}{\left(2\mathrm{y}-\mathrm{x}\right)}=\frac{\mathrm{x}}{\mathrm{z}} \\ \Rightarrow 2\mathrm{y}=\mathrm{x}+\mathrm{z} \\ \mathrm{Thus},x,y\mathrm{and}z\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \end{gathered}$$