a,b,c are in G.P. ⇒b2=ac
Discriminant of ax2+2bx+c=0 is 4b2−4ac=4ac−4ac=0 ⇒ roots are coincident and equals −2a2b=−ab
This must be the common root. Hence x=−ab must satisfy dxx2+2ex+f=0 ∴da2b2−2eab+f=0⇒db2−2eba+fa2=0⇒dac−2eba+fa2=0
dividing by a2c, ad−ac2eb+cf=0 ad−b2e+cf=0(b2=ac) ⇒ad,be,cfare inA.P