A) \[c\]
B) \[b\]
C) \[a\]
D) None of these
Correct Answer: B
Solution :
\[a+x,\ b+x,\ c+x\] are in H.P. \[\Rightarrow \] \[b+x=\frac{2(a+x)(c+x)}{(a+x)+(c+x)}\] \[\Rightarrow \] \[(b+x)(a+c+2x)=2(a+x)(c+x)\] \[\Rightarrow \] \[(a+c+2b)x+2{{x}^{2}}+ab+bc=2ac+2x(a+c)+2{{x}^{2}}\] \[\Rightarrow \] \[x(c+a-2b)=bc+ab-2ac\] \[\Rightarrow \] \[x(c+a-2b)=bc+ab-2{{b}^{2}}\] (\[\because \ a,\ b,\ c\] are in G.P.) Þ\[x(c+a-2b)=b(c+a-2b)\]Þ\[x=b\], if \[c+a-2b\ne 0\].
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