A) AP
B) GP
C) HP
D) None of these
Correct Answer: B
Solution :
Since,\[y-x,2(y-a)\]and\[(y-z)\]are in HP. \[\therefore \]\[2(y-a)=\frac{2(y-x)(y-z)}{y-x+y-z}\] \[\left[ \because b=\frac{2ac}{a+b} \right]\] \[\Rightarrow \]\[(2y-2a)(2y-x-z)=2(y-x)(y-z)\] \[\Rightarrow \]\[4{{y}^{2}}-2yx-2yz-4ay+2ax+2az\] \[=2{{y}^{2}}-2yz-2xy+2xz\] \[\Rightarrow \] \[2{{y}^{2}}-4ay+2ax+2az-2xz=0\] \[\Rightarrow \] \[{{y}^{2}}-2ay+ax+az-xz+{{a}^{2}}-{{a}^{2}}=0\] \[\Rightarrow \] \[{{y}^{2}}+{{a}^{2}}-2ay+a(x-a)-z(x-a)=0\] \[\Rightarrow \] \[{{(y-a)}^{2}}+(x-a)(a-z)=0\] \[\Rightarrow \] \[{{(y-a)}^{2}}=-(x-a)(a-z)\] \[\Rightarrow \] \[{{(y-a)}^{2}}=(x-a)(z-a)\] \[\therefore \]\[(x-a),(y-a),(z-a)\]are in GP.
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