A) \[2y=x+z\]
B) \[2x=y+z\]
C) \[2z=x+y\]
D) \[2xz=xy+yz\]
E) \[z=\frac{2xy}{x+y}\]
Correct Answer: D
Solution :
Since, \[x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}\] \[\therefore \] \[x=1+a+{{a}^{2}}+....\infty \] \[\Rightarrow \] \[x=\frac{1}{1-a}\] \[\Rightarrow \] \[(1-a)x=1\] \[\Rightarrow \] \[a=\frac{x-1}{x}\] Similarly,\[b=\frac{y-1}{y}\]and \[c=\frac{z-1}{z}\] Since, a, b and c are in AP. \[\therefore \] \[b=\frac{a+c}{2}\] \[\Rightarrow \] \[\frac{y-1}{y}=\frac{\frac{x-1}{x}+\frac{z-1}{z}}{2}\] \[\Rightarrow \] \[2xz(y-1)=y[z(x-1)+x(z-1)]\] \[\Rightarrow \] \[2xyz-2xz=xyz-yz+xyz-xy\] \[\Rightarrow \] \[-2xz=-yz-xy\] \[\Rightarrow \] \[2xz=xy+yz\]You need to login to perform this action.
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