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Question Number 175839 by Rasheed.Sindhi last updated on 08/Sep/22

If w, x, y and z be four consecutive  terms of any AP, then show that             w^2 −z^2 =3(x^2 −y^2 ).

$${If}\:{w},\:{x},\:{y}\:{and}\:{z}\:{be}\:{four}\:{consecutive} \\ $$$${terms}\:{of}\:{any}\:{AP},\:{then}\:{show}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{w}^{\mathrm{2}} −{z}^{\mathrm{2}} =\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right). \\ $$

Answered by mr W last updated on 08/Sep/22

let c=((x+y)/2), d=common difference  w=c−((3d)/2)  x=c−(d/2)  y=c+(d/2)  z=c+((3d)/2)  w^2 −z^2 =(c−((3d)/2))^2 −(c+((3d)/2))^2 =−6cd  x^2 −y^2 =(c−(d/2))^2 −(c+(d/2))^2 =−2cd  ⇒w^2 −z^2 =3(x^2 −y^2 )

$${let}\:{c}=\frac{{x}+{y}}{\mathrm{2}},\:{d}={common}\:{difference} \\ $$$${w}={c}−\frac{\mathrm{3}{d}}{\mathrm{2}} \\ $$$${x}={c}−\frac{{d}}{\mathrm{2}} \\ $$$${y}={c}+\frac{{d}}{\mathrm{2}} \\ $$$${z}={c}+\frac{\mathrm{3}{d}}{\mathrm{2}} \\ $$$${w}^{\mathrm{2}} −{z}^{\mathrm{2}} =\left({c}−\frac{\mathrm{3}{d}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({c}+\frac{\mathrm{3}{d}}{\mathrm{2}}\right)^{\mathrm{2}} =−\mathrm{6}{cd} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\left({c}−\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({c}+\frac{{d}}{\mathrm{2}}\right)^{\mathrm{2}} =−\mathrm{2}{cd} \\ $$$$\Rightarrow{w}^{\mathrm{2}} −{z}^{\mathrm{2}} =\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$

Commented by peter frank last updated on 08/Sep/22

thans

$$\mathrm{thans} \\ $$

Commented by Rasheed.Sindhi last updated on 08/Sep/22

Nice sir, thanks!

$$\mathbb{N}\mathrm{ice}\:\mathrm{sir},\:\mathrm{than}\Bbbk\mathrm{s}! \\ $$

Commented by Tawa11 last updated on 15/Sep/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Answered by Rasheed.Sindhi last updated on 08/Sep/22

x−w=y−x=z−y  x−w=y−x ∧ y−x=z−y  w=2x−y ,z=2y−x  w^2 −z^2 =(2x−y)^2 −(2y−x)^2            =4x^2 −4xy+y^2 −4y^2 +4xy−x^2            =3x^2 −3y^2 =3(x^2 −y^2 )

$${x}−{w}={y}−{x}={z}−{y} \\ $$$${x}−{w}={y}−{x}\:\wedge\:{y}−{x}={z}−{y} \\ $$$${w}=\mathrm{2}{x}−{y}\:,{z}=\mathrm{2}{y}−{x} \\ $$$${w}^{\mathrm{2}} −{z}^{\mathrm{2}} =\left(\mathrm{2}{x}−{y}\right)^{\mathrm{2}} −\left(\mathrm{2}{y}−{x}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{xy}+{y}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{4}{xy}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} =\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$

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