A) G.P
B) G.P. only if x<0
C) G.P. only if x>0
D) none of these
Correct Answer: A
Solution :
[a] \[\frac{2}{\sqrt{c}+\sqrt{a}}=\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{a}+\sqrt{b}}\] \[=\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{(\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b})}\] \[\Rightarrow \,\,\,2\sqrt{ab}+2b+2\sqrt{ac}+2\sqrt{bc}\] \[=2\sqrt{bc}+2\sqrt{ac}+c+2\sqrt{ab}+a\Rightarrow 2b=a+c\] \[\therefore \] a, b, c, are in A.P. \[\Rightarrow \,\,\,ax,bx,cx,\] are in A.P. \[\Rightarrow \,\,\,ax+1,\,bx+1,cx+1,\]are in A.P. \[\Rightarrow \,\,{{9}^{ax+1}},{{9}^{bx+1}},{{9}^{cx+1}}\]are in GP [See the properties of A.P & G.P.]
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