A) AP
B) GP
C) HP
D) None of these
Correct Answer: B
Solution :
It is given that \[y-x,2(y-a)\]and \[(y-z)\]are in HP. \[\Rightarrow \] \[\frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}\]are in AP. \[\Rightarrow \] \[\frac{1}{2(y-a)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2(y-a)}\] \[\Rightarrow \] \[\frac{2a-y-x}{y-x}\frac{y+z-2a}{y-z}\] \[\Rightarrow \] \[\frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)}\] \[\Rightarrow \] \[\frac{x-a}{y-a}=\frac{y-a}{z-a}\] Hence, \[x-a,y-a\]and \[z-a\] are in GP
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