Question

The roots of equation $a\left(b-c\right)x^2+b\left(c-a\right)x+c\left(a-b\right)=0$ are equal, then $a,b,c$ are in

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is C

H.P

Explanation for correct option

The given quadratic equation is $a\left(b-c\right)x^2+b\left(c-a\right)x+c\left(a-b\right)=0$

For the quadratic equation $p{x}^2+qx+r=0$ will have equal root when $q^2-4pr=0$

Now comparing with the given equation we have
$$\begin{gathered} {\left(b\left(c-a\right)\right)}^2-4·c\left(a-b\right)·a\left(b-c\right)=0 \\ \Rightarrow b^2\left(c^2-2ac+a^2\right)-4ac\left(ab-ac-b^2+bc\right)=0 \\ \Rightarrow b^2{c}^2-2ac{b}^2+a^2{b}^2-4a^{2}bc+4a^2{c}^2+4a{b}^{2}c-4ab{c}^2=0 \\ \Rightarrow {\left[b\left(a+c\right)-2ac\right]}^2=0 \\ \Rightarrow b\left(a+c\right)-2ac=0 \\ \Rightarrow ab+bc=2ac \\ \Rightarrow ab-ac=ac-bc \\ \Rightarrow \frac{1}{c}-\frac{1}{b}=\frac{1}{b}-\frac{1}{a} \\ \Rightarrow \frac{2}{b}=\frac{1}{c}+\frac{1}{a} \end{gathered}$$

If $a,b$ and $c$ are in harmonic progression then

$\frac{2}{b}=\frac{1}{c}+\frac{1}{a}$

Thus, a Harmonic Progression is formed.

Hence, the correct option is $\left(\mathrm{C}\right)$