Question

Let $p,q$ be roots of the equation $x^2-4x+A=0$ and $r$ and $s$ be the roots of the equation $x^2-20x+B=0$. If $p<q<r<s$ are in A.P., then $(A,B)$ is

• A. $(0,-96)$
• B. $(96,0)$
• C. $(0,96)$
• D. $(-96,0)$

Answer 1

Answer: (C)

$(0,96)$

$p,q,r,s$ are in A.P.
Let $p=a,q=a+d,r=a+2d,s=a+3d$
From the first equation, we have,
$a+a+d=4$
$\Rightarrow 2a+d=4$
From the second equation, we have
$2a+5d=20$.
On solving we get,
$a=0,d=4$
So, $p=0,q=4,r=8,s=12$
$A=pq=0$ , $B=rs=96$