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Question Number 25378 by Tinkutara last updated on 09/Dec/17

If log x, log y, log z (x,y,z > 1) are in GP then 2x+log(bx), 3x+log(by), 4x+log(bz) are in A.P. True/False?

$${If}\:\mathrm{log}\:{x},\:\mathrm{log}\:{y},\:\mathrm{log}\:{z}\:\left({x},{y},{z}\:>\:\mathrm{1}\right)\:{are}\:{in} \\ $$ $${GP}\:{then}\:\mathrm{2}{x}+\mathrm{log}\left({bx}\right),\:\mathrm{3}{x}+\mathrm{log}\left({by}\right), \\ $$ $$\mathrm{4}{x}+\mathrm{log}\left({bz}\right)\:{are}\:{in}\:{A}.{P}. \\ $$ $$\boldsymbol{{True}}/\boldsymbol{{False}}? \\ $$

Answered by Rasheed.Sindhi last updated on 09/Dec/17

((log y)/(log x))=((log z)/(log y)) Given To be proved: {3x+log(by)}−{2x+log(bx)} ={4x+log(bz)}−{3x+log(by)} x+log(by)−log(bx)=x+log(bz)−log(by) x+log(((by)/(bx)))=x+log(((bz)/(by))) log((y/x))=log((z/y)) log y−log x=log z−log y ∴ log x,log y, logz are in AP We can also prove vice versa i-e If log x,log y, logz are in AP then 2x+... , 3x+... , 4x+... are also in AP. ∴ If log x,log y, logz are in GP then 2x+..,3x+..,4x+... can′t be in GP (∵ a sequence can′t be both AP and GP simultneously in general.) Hence False.

$$\frac{\mathrm{log}\:{y}}{\mathrm{log}\:{x}}=\frac{\mathrm{log}\:{z}}{\mathrm{log}\:{y}}\:\:\:\mathrm{Given} \\ $$ $$\mathrm{To}\:\mathrm{be}\:\mathrm{proved}: \\ $$ $$\left\{\mathrm{3}{x}+\mathrm{log}\left({by}\right)\right\}−\left\{\mathrm{2}{x}+\mathrm{log}\left({bx}\right)\right\} \\ $$ $$\:\:\:\:\:=\left\{\mathrm{4}{x}+\mathrm{log}\left({bz}\right)\right\}−\left\{\mathrm{3}{x}+\mathrm{log}\left({by}\right)\right\} \\ $$ $${x}+\mathrm{log}\left({by}\right)−\mathrm{log}\left({bx}\right)={x}+\mathrm{log}\left({bz}\right)−\mathrm{log}\left({by}\right) \\ $$ $${x}+\mathrm{log}\left(\frac{{by}}{{bx}}\right)={x}+\mathrm{log}\left(\frac{{bz}}{{by}}\right) \\ $$ $$\mathrm{log}\left(\frac{{y}}{{x}}\right)=\mathrm{log}\left(\frac{{z}}{{y}}\right) \\ $$ $$\mathrm{log}\:\mathrm{y}−\mathrm{log}\:\mathrm{x}=\mathrm{log}\:\mathrm{z}−\mathrm{log}\:\mathrm{y}\:\: \\ $$ $$\therefore\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$ $$\mathrm{We}\:\mathrm{can}\:\mathrm{also}\:\mathrm{prove}\:\mathrm{vice}\:\mathrm{versa} \\ $$ $$\mathrm{i}-\mathrm{e}\:\mathrm{If}\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$ $$\mathrm{then}\:\mathrm{2x}+...\:,\:\mathrm{3x}+...\:,\:\mathrm{4x}+...\:\mathrm{are}\: \\ $$ $$\mathrm{also}\:\mathrm{in}\:\mathrm{AP}. \\ $$ $$\therefore\:\mathrm{If}\:\mathrm{log}\:\mathrm{x},\mathrm{log}\:\mathrm{y},\:\mathrm{logz}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP} \\ $$ $$\mathrm{then}\:\mathrm{2x}+..,\mathrm{3x}+..,\mathrm{4x}+...\:\mathrm{can}'\mathrm{t}\:\mathrm{be} \\ $$ $$\mathrm{in}\:\mathrm{GP}\:\left(\because\:\mathrm{a}\:\mathrm{sequence}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{both}\right. \\ $$ $$\left.\mathrm{AP}\:\mathrm{and}\:\mathrm{GP}\:\mathrm{simultneously}\:\mathrm{in}\:\mathrm{general}.\right) \\ $$ $$\mathrm{Hence}\:\mathrm{False}. \\ $$

Commented byTinkutara last updated on 09/Dec/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

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