Question

# If a, b, c are in G.P. and a1/x = b1/y = c1/z, then xyz are in (a) AP (b) GP (c) HP (d) none of these

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Solution

## (a) AP $a,b\mathrm{and}c\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}\therefore {b}^{2}=ac\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{log}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides}:\phantom{\rule{0ex}{0ex}}2\mathrm{log}\mathrm{b}=\mathrm{log}\mathrm{a}+\mathrm{log}\mathrm{c}........\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},{a}^{\frac{1}{x}}={b}^{\frac{1}{y}}={c}^{\frac{1}{z}}\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{log}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{side}s:\phantom{\rule{0ex}{0ex}}\frac{\mathrm{log}\mathrm{a}}{\mathrm{x}}=\frac{\mathrm{log}\mathrm{b}}{\mathrm{y}}=\frac{\mathrm{log}\mathrm{c}}{\mathrm{z}}........\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{comparing}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right):\phantom{\rule{0ex}{0ex}}\frac{\mathrm{log}a}{x}=\frac{\mathrm{log}\mathrm{a}+\mathrm{log}\mathrm{c}}{2y}=\frac{\mathrm{log}\mathrm{c}}{\mathrm{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{log}a}{x}=\frac{\mathrm{log}\mathrm{a}+\mathrm{log}\mathrm{c}}{2y}\mathrm{and}\frac{\mathrm{log}\mathrm{a}}{\mathrm{x}}=\frac{\mathrm{log}\mathrm{c}}{\mathrm{z}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{log}\mathrm{a}\left(2\mathrm{y}-\mathrm{x}\right)=\mathrm{xlog}\mathrm{c}\mathrm{and}\frac{\mathrm{log}\mathrm{a}}{\mathrm{log}\mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{log}\mathrm{a}}{\mathrm{log}\mathrm{c}}=\frac{\mathrm{x}}{\left(2\mathrm{y}-\mathrm{x}\right)}\mathrm{and}\frac{\mathrm{log}\mathrm{a}}{\mathrm{log}\mathrm{c}}=\frac{\mathrm{x}}{\mathrm{z}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{x}}{\left(2\mathrm{y}-\mathrm{x}\right)}=\frac{\mathrm{x}}{\mathrm{z}}\phantom{\rule{0ex}{0ex}}⇒2\mathrm{y}=\mathrm{x}+\mathrm{z}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},x,y\mathrm{and}z\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.\phantom{\rule{0ex}{0ex}}$

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