Question

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

• A. $A=-3$
• B. $B=-77$
• C. $B=77$
• D. $A=3$

Answer 1

Answer: (A), (C)

$B=77$

Since $p,q$ are the roots of the equation $x^2-2x+A=0$
$\therefore p+q=2,pq=A$

$r,s$ are the roots of the equation $x^2-18x+B=0$
$\therefore r+s=18,rs=B$

It is given that $p,q,r,s$ are in A.P.
Therefore, let $p=a,q=a+d,r=a+2d,s=a+3d$
Hence, $a+(a+d)=2$
and $(a+2d)+(a+3d)=18$
$\Rightarrow 2a+d=2\cdots \left(1\right)$
and $2a+5d=18\cdots \left(2\right)$
Solving eqns $\left(1\right)$ and $\left(2\right)$, we get
$a=-1$ and $d=4$
$\therefore p=-1,q=3,r=7,s=11$
Hence, $A=pq=-3$ and $B=rs=77$