∵p,q,r,s are in AP. Let p=a,q=a+d,r=a+2d,s=a+3d Given, p.q are roots of x2−2x+A=0 Then, p+q=2 and pq=A ⇒2a+d=2 ...(i) Also, r and s are roots of x2−18x+B=0 Then,r+s=18andrs=B ⇒2a+5d=18 ...(ii) Solving Eqs. (i) and (ii), we get a=−1,d=4 Now, pq=A (product of roots) OrA=a(a+d)=(−1)(−1+4) A=−3 andrs=B or B=(a+2d)(a+3d)1=(−1+8)(−1+12)=77 ∴A=−3,B=77