Question

Lets consider quadratic equation $a{x}^2+bx+c=0$ where $a,b,c\in R$ and $a\ne 0$. If above equation has roots $\alpha ,\beta$, then $\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$ and the equation can be written as $a{x}^2+bx+c=a(x-\alpha )(x-\beta )$. Also, if $a_1,a_2,a_3,a_4$, ..... are in A.P., then $a_2-a_1=a_3-a_2=a_4-a_3=...\ne 0$ and if $b_1,b_2,b_3,b_4$, ... are in G.P., then $\frac{b_2}{b_1}=\frac{b_3}{b_2}=\frac{b_4}{b_3}=$ ... $\ne 1$ Now if $c_1$, $c_2$, $c_3$, $c_4$, ... are in HP, then $\frac{1}{c_2}-\frac{1}{c_1}=\frac{1}{c_3}-\frac{1}{c_2}=\frac{1}{c_4}-\frac{1}{c_3}=$... $\ne 0$. If the roots of equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal, then $a,b,c$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $\text{None of these}$

Answer 1

Answer: (C)

$H.P.$

Given quadratic equation

$a(b-c)x^2+b(c-a)x+c(a-b)=0$

Here, $a(b-c)+b(c-a)+c(a-b)=0$

So, one root is $1$

Let the other root be $\alpha$

$1.\alpha =\frac{c(a-b)}{a(b-c)}\dots \left(productofroots\right)$

$\Rightarrow \alpha =\frac{c(a-b)}{a(b-c)}$

Given the roots are equal, then

$1=\frac{c(a-b)}{a(b-c)}$

$\Rightarrow \frac{a-b}{b-c}=\frac{a}{c}$

$\Rightarrow b=\frac{2ac}{a+c}$

So, $a,b,c$ are in $H.P.$