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Question-195944




Question Number 195944 by sonukgindia last updated on 13/Aug/23
Answered by AST last updated on 13/Aug/23
a_2 −a_1 =a_3 −a_2 ⇒xy−(x/y)=x+y−xy  a_4 −a_3 =a_3 −a_2 ⇒−2y=x+y−xy  ⇒−2y=xy−(x/y)=x(y−(1/y))⇒x=((−2y^2 )/(y^2 −1))...(i)  a_5 =a_4 +d=(x−y)+(x+y−xy)=2x−xy  a_5 =(x−y)+xy−(x/y)=2x−xy⇒  ⇒x−xy−xy+(x/y)=−y⇒x(1−2y+(1/y))=−y  ⇒x=((−y)/((y−2y^2 +1)/y))=((−y^2 )/(y−2y^2 +1))...(ii)  (i)=(ii)⇒((−y^2 )/(y−2y^2 +1))=((−2y^2 )/(y^2 −1))⇒y^2 −1=2y−4y^2 +2  ⇒5y^2 −2y−3=0⇒y=1 or ((−3)/5)  y=−(3/5)⇒x=((−(9/(25)))/((−15−18+25)/(25)))=((−9)/(−8))=(9/8)  ⇒a_1 =((−15)/8),a_2 =((−27)/(40)),a_3 =((21)/(40)),a_4 =((69)/(40)),a_5 =((117)/(40))
$${a}_{\mathrm{2}} −{a}_{\mathrm{1}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \Rightarrow{xy}−\frac{{x}}{{y}}={x}+{y}−{xy} \\ $$$${a}_{\mathrm{4}} −{a}_{\mathrm{3}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \Rightarrow−\mathrm{2}{y}={x}+{y}−{xy} \\ $$$$\Rightarrow−\mathrm{2}{y}={xy}−\frac{{x}}{{y}}={x}\left({y}−\frac{\mathrm{1}}{{y}}\right)\Rightarrow{x}=\frac{−\mathrm{2}{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −\mathrm{1}}…\left({i}\right) \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{4}} +{d}=\left({x}−{y}\right)+\left({x}+{y}−{xy}\right)=\mathrm{2}{x}−{xy} \\ $$$${a}_{\mathrm{5}} =\left({x}−{y}\right)+{xy}−\frac{{x}}{{y}}=\mathrm{2}{x}−{xy}\Rightarrow \\ $$$$\Rightarrow{x}−{xy}−{xy}+\frac{{x}}{{y}}=−{y}\Rightarrow{x}\left(\mathrm{1}−\mathrm{2}{y}+\frac{\mathrm{1}}{{y}}\right)=−{y} \\ $$$$\Rightarrow{x}=\frac{−{y}}{\frac{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}{{y}}}=\frac{−{y}^{\mathrm{2}} }{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}…\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right)\Rightarrow\frac{−{y}^{\mathrm{2}} }{{y}−\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}}=\frac{−\mathrm{2}{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} −\mathrm{1}}\Rightarrow{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{2}{y}−\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Rightarrow\mathrm{5}{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{3}=\mathrm{0}\Rightarrow{y}=\mathrm{1}\:{or}\:\frac{−\mathrm{3}}{\mathrm{5}} \\ $$$${y}=−\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{x}=\frac{−\frac{\mathrm{9}}{\mathrm{25}}}{\frac{−\mathrm{15}−\mathrm{18}+\mathrm{25}}{\mathrm{25}}}=\frac{−\mathrm{9}}{−\mathrm{8}}=\frac{\mathrm{9}}{\mathrm{8}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} =\frac{−\mathrm{15}}{\mathrm{8}},{a}_{\mathrm{2}} =\frac{−\mathrm{27}}{\mathrm{40}},{a}_{\mathrm{3}} =\frac{\mathrm{21}}{\mathrm{40}},{a}_{\mathrm{4}} =\frac{\mathrm{69}}{\mathrm{40}},{a}_{\mathrm{5}} =\frac{\mathrm{117}}{\mathrm{40}} \\ $$

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