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 $A$ = _____________.

Answer 1

Let p and q be roots of the equation $x^2-2x+A=0$.

Then,

$p+q=2,pq=A$

Let r and s be roots of the equation $x^2-18x+B=0$.

Then,

$r+s=18,rs=B$

And it is given that p, q, r, and s are in A.P. Let $p=a-3b,q=a-d,r=a+d$ and $s=a+3d$. As $p<q<r<s$, we have $d>0$. Now,

$2=p+q=a-3d+a-d=2a-4d$

$\to a-2d=1$ (1)

and

$18=r+s=a+d+a+3d$

$\Rightarrow a+2d=9$ (2)

Solving (1) and (2), $a=5,d=2$

$\therefore$ $p=-1,q=3,r=7,s=11$

Therefore, $A=pq=-3andB=rs=77.$