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 find the value of $A+B$.

Answer 1

$x^2-2x+A=0p+q=2pq=A{x}^2-16x+B=0r+s=16rs=B$

Also, p<q<r<s are in A.P.

Therefore, q-p=r-q=s-r

$⟹2q=r+p⟹2r=q+s⟹q=\frac{r+p}{2}⟹r=\frac{q+s}{2}⟹2q-p=r⟹s=2r-qr+s=18⟹2q-p+2r-q=18⟹q-p+2r-q=18⟹q-p+4q-2p=18⟹5q-3p=18⟹5(2-p)-2p=18⟹10-5p-3p=18⟹-8p=8⟹p=-1q=2-p=2+1=3\therefore A=pq=(-1)\times 3=-3r=2q-p=6+1=7s=2r-q=14-3=11\therefore rs=77\therefore A+B=pq+rs=-3+77=74$