A) \[2b=a+c\]
B) \[2a=b+c\]
C) \[2c=a+b\]
D) \[\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\]
Correct Answer: B
Solution :
Since the given equation has equal roots, \[\therefore \] \[D=0\] \[\Rightarrow \] \[{{(b-c)}^{2}}-4(c-a)\,(a-b)=0\] \[\Rightarrow \] \[4{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-4ab+2bc-4ac=0\] \[\Rightarrow \] \[{{(-2a)}^{2}}+({{b}^{2}})+{{(c)}^{2}}+2(-2a)(b)+2bc\] \[+2(-2a)\,(c)=0\] \[\Rightarrow \] \[{{(-2a+b+c)}^{2}}=0\] \[\Rightarrow \] \[-2a+b+c=0\] \[\Rightarrow \]\[2a=b+c\]
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