Question Number 195944 by sonukgindia last updated on 13/Aug/23
Answered by AST last updated on 13/Aug/23
$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \Rightarrow{xy}−\frac{{x}}{{y}}={x}+{y}−{xy} \\ $$$${a}_{\mathrm{4}} −{a}_{\mathrm{3}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \Rightarrow−\mathrm{2}{y}={x}+{y}−{xy} \\ $$$$\Rightarrow−\mathrm{2}{y}={xy}−\frac{{x}}{{y}}={x}\left({y}−\frac{\mathrm{1}}{{y}}\right)\Rightarrow{x}=\frac{−\mathrm{2}{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −\mathrm{1}}…\left({i}\right) \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} +{d}=\left({x}−{y}\right)+\left({x}+{y}−{xy}\right)=\mathrm{2}{x}−{xy} \\ $$$${a}_{\mathrm{5}} =\left({x}−{y}\right)+{xy}−\frac{{x}}{{y}}=\mathrm{2}{x}−{xy}\Rightarrow \\ $$$$\Rightarrow{x}−{xy}−{xy}+\frac{{x}}{{y}}=−{y}\Rightarrow{x}\left(\mathrm{1}−\mathrm{2}{y}+\frac{\mathrm{1}}{{y}}\right)=−{y} \\ $$$$\Rightarrow{x}=\frac{−{y}}{\frac{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{{y}}}=\frac{−{y}^{\mathrm{2}} }{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}…\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right)\Rightarrow\frac{−{y}^{\mathrm{2}} }{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}=\frac{−\mathrm{2}{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −\mathrm{1}}\Rightarrow{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{2}{y}−\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Rightarrow\mathrm{5}{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{3}=\mathrm{0}\Rightarrow{y}=\mathrm{1}\:{or}\:\frac{−\mathrm{3}}{\mathrm{5}} \\ $$$${y}=−\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{x}=\frac{−\frac{\mathrm{9}}{\mathrm{25}}}{\frac{−\mathrm{15}−\mathrm{18}+\mathrm{25}}{\mathrm{25}}}=\frac{−\mathrm{9}}{−\mathrm{8}}=\frac{\mathrm{9}}{\mathrm{8}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{−\mathrm{15}}{\mathrm{8}},{a}_{\mathrm{2}} =\frac{−\mathrm{27}}{\mathrm{40}},{a}_{\mathrm{3}} =\frac{\mathrm{21}}{\mathrm{40}},{a}_{\mathrm{4}} =\frac{\mathrm{69}}{\mathrm{40}},{a}_{\mathrm{5}} =\frac{\mathrm{117}}{\mathrm{40}} \\ $$