A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
If \[\frac{x+y}{2},\ y,\ \frac{y+z}{2}\] are in H.P., then \[=\frac{1}{2}{{x}_{1}}{{y}_{1}}\left| \ \begin{matrix} 1 & 1 & 1 \\ r & r & 1 \\ {{r}^{2}} & {{r}^{2}} & 1 \\ \end{matrix}\ \right|=0\] \[y=\frac{xy+xz+{{y}^{2}}+yz}{x+2y+z}\] \[\Rightarrow \] \[xy+2{{y}^{2}}+yz=xy+xz+{{y}^{2}}+yz\]\[\Rightarrow \]\[{{y}^{2}}=xz\] Thus \[x,\ y,\ z\] will be in G.P.
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