Question

Let $A=a^{2}b+a{b}^2-a^{2}c-a{c}^2,$ $B=b^{2}c+b{c}^2-a^{2}b-a{b}^2$ and $C=a^{2}c+a{c}^2-b^{2}c-b{c}^2$, where $a>b>c>0$. If the equation $A{x}^2+Bx+C=0$ has equal roots, then $a,b,c$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. A.G.P.

Answer 1

The correct option is C H.P.

Given : $A=a^{2}b+a{b}^2-a^{2}c-a{c}^2,$ $B=b^{2}c+b{c}^2-a^{2}b-a{b}^2$ and $C=a^{2}c+a{c}^2-b^{2}c-b{c}^2$
By observation, we see
$A+B+C=0$
Now, the given equation is
$A{x}^2+Bx+C=0$
So, one root of the equation is $1$, as both roots are equal so both root are equal to $1$,
Product of roots
$\frac{C}{A}=1\Rightarrow A=C\Rightarrow a^{2}b+a{b}^2-a^{2}c-a{c}^2=a^{2}c+a{c}^2-b^{2}c-b{c}^2\Rightarrow a\left[a\right(b-c)+(b^2-c^2\left)\right]=c\left[c\right(a-b)+(a^2-b^2\left)\right]\Rightarrow a(b-c)(a+b+c)=c(a-b)(a+b+c)\Rightarrow ab-ac=ac-bc\Rightarrow b(a+c)=2ac\therefore b=\frac{2ac}{a+c}$

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