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If $\large\frac{1}{x+y}$, $\large\frac{1}{xy}$, $\large\frac{1}{y+z}$ are in AP, $x, y, z$ are in

$(a)\;AP\qquad(b)\;GP\qquad(c)\;HP\qquad(d)\;None\;of\;the\;above$

1 Answer

Answer : (b) GP
Explanation : $\;\large\frac{1}{x+y}\;,\;\large\frac{1}{xy}\;,\;\large\frac{1}{y+z}\;in\;AP$
$2\;(1/2y)\;=\large\frac{1}{x+y}+\frac{1}{y+z}$
$\large\frac{1}{y}=\large\frac{y+z+x+y}{(x+y)(y+z)}$
$(x+y)(y+z)=y(2y+x+z)$
$xy+y^2+xz+yz=2y^2+xy+yz$
$y^2+xz=2y^2$
$xz=y^2$
$x,y,z\;are \;in\;AP.$
answered Jan 3, 2014 by yamini.v
edited Jan 24, 2014
 

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