Question

Let $p$ and $q$ be roots of the equation $x^2-2x+A=0$ and let $r$ and $s$ be the roots of the equation $x^2-18x+B=0$. If $p<q<r<s$ are in arithmetic progression, then

• A. $A=-83,B=-3$
• B. $A=-3,B=77$
• C. $q=3,r=7$
• D. $p+q+r+s=20$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $A=-3,B=77$
C $q=3,r=7$
D $p+q+r+s=20$

Given $p$ and $q$ be roots of the equation $x^2-2x+A=0$ and $r$ and $s$ be the roots of the equation $x^2-18x+B=0$

The standard quadratic equation is $a{x}^2+bx+c=0$

Then Sum of roots = $-\frac{b}{a}$

and Product of roots = $\frac{c}{a}$

$\therefore p+q=2,pq=A$

$r+s=18,rs=B$

Given that $p,q,r,s$ are in A.M.

So, $\frac{r+p}{2}=q,\frac{q+s}{2}=r$....(1)

It is clear that $p+q+r+s=20$.

In equation (1), all variables converted in $p,r$ by using the above equation.

$4-3p=r,3r+p=20$

So, we get $r=7,p=-1,q=3,s=11$.

$\Rightarrow A=-3,B=77$.

Hence, options 'B', 'C' and 'D' are correct.