Question

If the sum of the roots of the quadratic equation $a x^{2}+$ $b x+c=0$ is equal to the sum of the squares of their reciprocals, then $\frac{a}{c}, \frac{b}{a}$ and $\frac{c}{b}$ are in

• A. arithmetic progression
• B. geometric progression
• C. harmonic progression
• D. arithmetico-geometric progression

Answer 1

Answer: (C)

Given equation is
$a x^{2}+b x+c=0$
Let $\alpha, \beta$ be the roots of this equation.
then, $\alpha+\beta=-\frac{b}{a}$ and $a \beta=\frac{c}{a}$
Also, $\alpha+\beta=\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\frac{a^{2}+\beta^{2}}{\alpha^{2} \beta^{2}}$
$=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{(\alpha \beta)^{2}}$
$\Rightarrow \alpha+\beta=\left(\frac{\alpha+\beta}{\alpha \beta}\right)^{2}-\frac{2}{\alpha \beta}$
$\Rightarrow \left(-\frac{b}{a}\right)=\left(\frac{-b / a}{c / a}\right)^{2}-\frac{2}{c / a} $
$\Rightarrow -\frac{b}{a}=\left(\frac{b}{c}\right)^{2}-\frac{2 a}{c}$
$\Rightarrow \frac{2 a}{c}=\left(\frac{b}{c}\right)^{2}+\frac{b}{a} $
$\Rightarrow \frac{2 a}{c}=\frac{b}{c}\left[\frac{b}{c}+\frac{c}{a}\right]$
$\Rightarrow \frac{2 a}{b}=\frac{b}{c}+\frac{c}{a} $
$\Rightarrow \frac{c}{a}, \frac{a}{b}, \frac{b}{c}$ are in A.P.
$\Rightarrow \frac{a}{c}, \frac{b}{a}, \frac{c}{b}$ are in H.P.