Question Number 25378 by Tinkutara last updated on 09/Dec/17
$${If}\:\mathrm{log}\:{x},\:\mathrm{log}\:{y},\:\mathrm{log}\:{z}\:\left({x},{y},{z}\:>\:\mathrm{1}\right)\:{are}\:{in} \\ $$$${GP}\:{then}\:\mathrm{2}{x}+\mathrm{log}\left({bx}\right),\:\mathrm{3}{x}+\mathrm{log}\left({by}\right), \\ $$$$\mathrm{4}{x}+\mathrm{log}\left({bz}\right)\:{are}\:{in}\:{A}.{P}. \\ $$$$\boldsymbol{{True}}/\boldsymbol{{False}}? \\ $$
Answered by Rasheed.Sindhi last updated on 09/Dec/17
$$\frac{\mathrm{log}\:{y}}{\mathrm{log}\:{x}}=\frac{\mathrm{log}\:{z}}{\mathrm{log}\:{y}}\:\:\:\mathrm{Given} \\ $$$$\mathrm{To}\:\mathrm{be}\:\mathrm{proved}: \\ $$$$\left\{\mathrm{3}{x}+\mathrm{log}\left({by}\right)\right\}−\left\{\mathrm{2}{x}+\mathrm{log}\left({bx}\right)\right\} \\ $$$$\:\:\:\:\:=\left\{\mathrm{4}{x}+\mathrm{log}\left({bz}\right)\right\}−\left\{\mathrm{3}{x}+\mathrm{log}\left({by}\right)\right\} \\ $$$${x}+\mathrm{log}\left({by}\right)−\mathrm{log}\left({bx}\right)={x}+\mathrm{log}\left({bz}\right)−\mathrm{log}\left({by}\right) \\ $$$${x}+\mathrm{log}\left(\frac{{by}}{{bx}}\right)={x}+\mathrm{log}\left(\frac{{bz}}{{by}}\right) \\ $$$$\mathrm{log}\left(\frac{{y}}{{x}}\right)=\mathrm{log}\left(\frac{{z}}{{y}}\right) \\ $$$$\mathrm{log}\:\mathrm{y}−\mathrm{log}\:\mathrm{x}=\mathrm{log}\:\mathrm{z}−\mathrm{log}\:\mathrm{y}\:\: \\ $$$$\therefore\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{also}\:\mathrm{prove}\:\mathrm{vice}\:\mathrm{versa} \\ $$$$\mathrm{i}-\mathrm{e}\:\mathrm{If}\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$$$\mathrm{then}\:\mathrm{2x}+…\:,\:\mathrm{3x}+…\:,\:\mathrm{4x}+…\:\mathrm{are}\: \\ $$$$\mathrm{also}\:\mathrm{in}\:\mathrm{AP}. \\ $$$$\therefore\:\mathrm{If}\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP} \\ $$$$\mathrm{then}\:\mathrm{2x}+..,\mathrm{3x}+..,\mathrm{4x}+…\:\mathrm{can}'\mathrm{t}\:\mathrm{be} \\ $$$$\mathrm{in}\:\mathrm{GP}\:\left(\because\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{both}\right. \\ $$$$\left.\mathrm{AP}\:\mathrm{and}\:\mathrm{GP}\:\mathrm{simultneously}\:\mathrm{in}\:\mathrm{general}.\right) \\ $$$$\mathrm{Hence}\:\mathrm{False}. \\ $$
Commented by Tinkutara last updated on 09/Dec/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$