Question

Let $p$ and $q$ be the 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 $A=$ ... and $B=$ ....

Answer 1

Given, $p+q=2,pq=A$
and $r+s=18,rs=B$
and it is given that $p,q,r,s$ are in an AP.
Therefore, let $p=a-3d,q=a-d,r=a+d$
and $s=a+3d$
Since, $p<q<r<s$
We have, $d>0$
Now, $2=p+q=a-3d+a-d=2a-4d$
$\Rightarrow a-2d=1$ ...(i)
Again, $18=r+s=a+d+a+3d$
$18=2a+4d$
$\Rightarrow 9=a+2d$ ...(ii)
On substracting Eq. (i) from Eq. (ii). we get
$8=4d$
$\Rightarrow d=2$
On putting in Eq. (ii), we get $a=5$
$\therefore p=a-3d=5-6=-1$
$q=a-d=5-2=3$
$r=a+d=5+2=7$
and $s=a+3d=5+6=11$
Therefore, $A=pq=-3$ and $B=rs=77$