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 A.P., then $\frac{B-A}{10}$ is

Answer 1

Let $p,q,r,s$ be $a-3d,a-d,a+d,a+3d$ respectively,
Thus, as per the first equation $x^2-2x+A=0$
Sum of roots $p+q=2$
Therefore, $2a-4d=2$
$a-2d=1$-------------(1)
For the other equation, $x^2-18x+B=0$
$r+s=18$, $2a+4d=18$
$a+2d=9$--------------(2)
On adding (1) and (2),
$2a=10,a=5$
$d=2$
Now, $A=pq=(a-3d)(a-d)=-3$
$B=rs=(a+d)(a+3d)=7\times 11=77$
$\frac{B-A}{10}=\frac{77-(-3)}{10}$
$=8$