Question

If $a,b,c$ are in $A.P$ and if $(b-c)x^2+(c-a)+a-b=0$ and $2(c+a)x^2+(b+c)x=0$ have a common root then

• A. $a^2,b^2,c^2$ are in A.P
• B. $a^2,c^2,b^2$ are in A.P
• C. $a^2,c^2,b^2$ are in G.P
• D. $a^2,b^2,c^2$ are in H.P

Answer 1

The correct option is A $a^2,b^2,c^2$ are in A.P

Given a,b,c are in AP

$2b=a+c$

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

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

$(b-c)x(x-1)-(a-b)(x-1)=0$

$(x-1)\left[\right(b-c)x-(a-b\left)\right]=0$

$x=\frac{a-b}{b-c},1$

$x=\frac{a-b}{a-b}(\because 2b=a+cb-a=c-b)$

$x=1$

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

$x\left[2\right(c+a)x+(b+c\left)\right]=0$

$x=0$

$x=\frac{-(b+c)}{2.2b}$

$x=-\frac{b+c}{4b}$

Now

$\frac{-(b+c)}{4b}=1$

$-4b=b+c$

$5b=-c$

$a+c=2b$

$a-5b=2b$

$a=7d$

$c^2-a^2=25b^2-49b^2=-24b^2$

$b^2-c^2=b^2-25b^2=-24b^2$

$So,$

$c^2-a^2=b^2-c^2$

hence $a^2,c^2,b^2$ are in Ap