If $a, b, c$ are in $G.P.$ and $q^{\frac{1}{x}}=k^{\frac{1}{y}}=c^{\frac{1}{2}},$ prove that $x, y, z$ are in $A.P.$
If $a, b, c$ are in $G.P.$ and $q^{\frac{1}{x}}=k^{\frac{1}{y}}=c^{\frac{1}{2}},$ prove that $x, y, z$ are in $A.P.$
Let $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$ Then
$a=k^{x}, b=k^{y}$ and $c=k^{z}$ .........$(1)$
Since $a, b, c$ are in $G.P.,$ therefore,
$b^{2}=a c$ ..........$(2)$
Using $(1)$ in $(2),$ we get
$k^{2 y}=k^{x+z},$ which gives $2 y=x+z$
Hence, $x, y$ and $z$ are in $A.P.$
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