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. Þ \[\frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}\] are in A.P. Þ \[\frac{1}{2(y-a)}-\frac{1}{(y-x)}=\frac{1}{y-z}-\frac{1}{2(y-a)}\] Þ \[\frac{y-x-2y+2a}{(y-x)}=\frac{2y-2a-y+z}{(y-a)-(z-a)}\] \[\Rightarrow \frac{-x-y+2a}{(y-x)}=\frac{y+z-2a}{(y-z)}\] \[\Rightarrow \frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)}\] Þ \[\frac{(x-a)}{(y-a)}=\frac{(y-a)}{(z-a)}\] \[(x-a),(y-a),(z-a)\] are in G.P.
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