A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: B
Solution :
\[y-x,\ 2(y-a),\ (y-z)\] are in H.P. \[\Rightarrow \] \[\frac{1}{y-x},\ \frac{1}{2(y-a)},\ \frac{1}{y-z}\]are in A.P. \[\Rightarrow \] \[\frac{1}{2(y-a)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2(y-a)}\] \[\Rightarrow \] \[\frac{2a-y-x}{y-x}=\frac{y+z-2a}{y-z}\] \[\Rightarrow \] \[\therefore \] \[\Rightarrow \] \[\frac{x-a}{y-a}=\frac{y-a}{z-a}\] (Applying componendo and dividendo) \[\Rightarrow \] \[x-a,\ y-a,\ z-a\] are in G.P.
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