Given that α,γ are roots of the equation Ax2−4x+1=0 and β,δ the roots of the equation of Bx2−6x+1=0,such that α,β,γ and δ are in H.P., then

Since, α,β,γ,δ are in H.P., 1/α,1/β,1/γ,1/δ are in A.P. and they may be taken as a−3d,a−d,a+d,a+3d.Replacing x by 1/x, we get the equation whose roots are 1/α,1/β,1/γ,1/δ. Therefore,  equation x2−4x+A=0 has roots a−d,a+3dand equation2(a−d)=4,2(a+d)=6∴ a=5/2,d=1/2Product of the roots is(a−3d)(a+d)=A=3(a−d)(a+3d)=B=8