Question-30780

Question Number 30780 by mondodotto@gmail.com last updated on 25/Feb/18

Answered by Rasheed.Sindhi last updated on 25/Feb/18

(4x,4y,4z are in GP⇒x,y,z are in GP because ((4y)/(4x))=((4z)/(4y))⇒(y/x)=(z/y)) y=x+d ,z=x+2d x+(x+d)+(x+2d)=70 3x+3d=70⇒x+d=((70)/3)⇒x=((70−3d)/3) ((x+d)/x)=((x+2d)/(x+d)) (x+d)^2 =x(x+2d) (((70)/3))^2 =(((70−3d)/3))(((70−3d)/3)+2d) (((70)/3))^2 =(((70−3d)/3))(((70+3d)/3)) 70^2 −(3d)^2 =70^2 d=0 x=((70−3(0))/3)=((70)/3) x=y=z=((70)/3) common difference=0 common ratio=1

$$\left(\mathrm{4x},\mathrm{4y},\mathrm{4z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\Rightarrow\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{GP}\right. \\ $$$$\left.\mathrm{because}\:\frac{\mathrm{4y}}{\mathrm{4x}}=\frac{\mathrm{4z}}{\mathrm{4y}}\Rightarrow\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}}\right) \\ $$$$\mathrm{y}=\mathrm{x}+\mathrm{d}\:,\mathrm{z}=\mathrm{x}+\mathrm{2d} \\ $$$$\mathrm{x}+\left(\mathrm{x}+\mathrm{d}\right)+\left(\mathrm{x}+\mathrm{2d}\right)=\mathrm{70} \\ $$$$\mathrm{3x}+\mathrm{3d}=\mathrm{70}\Rightarrow\mathrm{x}+\mathrm{d}=\frac{\mathrm{70}}{\mathrm{3}}\Rightarrow\mathrm{x}=\frac{\mathrm{70}−\mathrm{3d}}{\mathrm{3}} \\ $$$$\frac{\mathrm{x}+\mathrm{d}}{\mathrm{x}}=\frac{\mathrm{x}+\mathrm{2d}}{\mathrm{x}+\mathrm{d}} \\ $$$$\left(\mathrm{x}+\mathrm{d}\right)^{\mathrm{2}} =\mathrm{x}\left(\mathrm{x}+\mathrm{2d}\right) \\ $$$$\left(\frac{\mathrm{70}}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{70}−\mathrm{3d}}{\mathrm{3}}\right)\left(\frac{\mathrm{70}−\mathrm{3d}}{\mathrm{3}}+\mathrm{2d}\right) \\ $$$$\left(\frac{\mathrm{70}}{\mathrm{3}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{70}−\mathrm{3d}}{\mathrm{3}}\right)\left(\frac{\mathrm{70}+\mathrm{3d}}{\mathrm{3}}\right) \\ $$$$\mathrm{70}^{\mathrm{2}} −\left(\mathrm{3d}\right)^{\mathrm{2}} =\mathrm{70}^{\mathrm{2}} \\ $$$$\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{70}−\mathrm{3}\left(\mathrm{0}\right)}{\mathrm{3}}=\frac{\mathrm{70}}{\mathrm{3}} \\ $$$$\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{\mathrm{70}}{\mathrm{3}} \\ $$$$\mathrm{common}\:\mathrm{difference}=\mathrm{0} \\ $$$$\mathrm{common}\:\mathrm{ratio}=\mathrm{1} \\ $$$$ \\ $$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

$$\mathrm{sir}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP}\: \\ $$

Commented by Rasheed.Sindhi last updated on 25/Feb/18

Yes sir, that′s why I take y=x+d & z=x+2d

$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{that}'\mathrm{s}\:\mathrm{why}\:\mathrm{I}\:\mathrm{take}\:\mathrm{y}=\mathrm{x}+\mathrm{d}\:\&\:\mathrm{z}=\mathrm{x}+\mathrm{2d} \\ $$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

$$\mathrm{thanx} \\ $$

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