Question

If x,y,z are in A.p.,
then find the value of (x+y-z) (y+z-x)

Answer 1

HEY MATE,
please see my answer
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x,y,z are inAP

so,

$=> 2y = x+z. -(1)$

$now, => (x+y-z)(y+z-x) => xy + xz - {x}^{2} +{y}^{2} + zy - yx - zy -{z}^{2} +zx )$

$= > (2zx - {x }^{2} + {y}^{2} - {z}^{2} )$

$= > - ( {x}^{2} + {z}^{2} - 2xz - {y}^{2} )$

$= > - ( {(x + z)}^{2} - {y}^{2} )$

$= > -( ( {2y)}^{2} - {y)}^{2}$

by eq. (1)

$= > - (4 {y}^{2} - {y}^{2} )$
$= > - 3 {y}^{2}$
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