A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
Since the roots of the quadratic equation in \[x\] are equal, we have \[({{B}^{2}}-4AC=0)\] \[\Rightarrow \] \[{{b}^{2}}{{(c-a)}^{2}}-4ac(b-c)(a-b)=0\] \[\Rightarrow \]\[{{b}^{2}}({{c}^{2}}-2ca+{{a}^{2}})-4ac(ba-{{b}^{2}}-ca+bc)=0\] \[\Rightarrow \]\[{{b}^{2}}({{c}^{2}}+2ca+{{a}^{2}})-4ac\left\{ b(a+c)-ac \right\}=0\] \[\Rightarrow \]\[{{b}^{2}}{{(a+c)}^{2}}-4ac\{b(a+c)-ac\}=0\] which can be seen to be true, if \[b=\frac{2ac}{a+c}\] or \[b(a+c)=2ac\] \[i.e.\], if \[a,\ b,\ c\]are in H.P.
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