Question

If $A,G$ and $H$ are respectively arithmetic, geometric and harmonic means between $a$ and $b$ both being unequal and positive, then $A=\frac{a+b}{2}\Rightarrow a+b=2A,G=\sqrt{ab}\Rightarrow ab=G^2$ and $H=\frac{2ab}{a+b}\Rightarrow G^2=AH$
From the above discussion we can say that $a,b$ are the roots of the equation $x^2-2Ax+G^2=0$
Now,quadratic equation, $x^2-Px+Q=0$ and quadratic equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ have a root common and satisfy the relation $b=\frac{2ac}{a+c},$ where $a,b,c$ are real numbers.
On the basis of the above information, answer the following questions:

The value of $\frac{a(b-c)}{b(c-a)}$:

• A. $-2$
• B. $2$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$-\frac{1}{2}$

Given$:a(b-c)x^2+b(c-a)x+c(a-b)=0$ ......$\left(1\right)$

$\because a(b-c)+b(c-a)+c(a-b)=0$
$\therefore x=1$ is a root of $\left(1\right)$

Let the other root be $\alpha$ then

$1\times \alpha =\frac{c(a-b)}{a(b-c)}=\frac{c(a-\frac{2ac}{a+c})}{a(\frac{2ac}{a+c}-c)}$
$=\frac{c{a}^2+c^{2}a-2a{c}^2}{2a^{2}c-a^{2}c-a{c}^2}=\frac{c{a}^2-a{c}^2}{a^{2}c-a{c}^2}=1$

$\therefore \alpha =1$

Hence, both the roots of eqn$\left(1\right)$ are $1,1$

then sum of the roots$=1+1=-\frac{b(c-a)}{a(b-c)}$
or $\frac{b(c-a)}{a(b-c)}=-2$

or $\frac{a(b-c)}{b(c-a)}=-\frac{1}{2}$