∵a,b,c are in G.P. ⇒b2=ac …(i)
Equation ax2+2bx+c=0
Hence, D=4b2−4ac=0 (∵b2=ac)
⇒ equation has equal roots ⇒ roots are =a−b,a−b ( ∵ sum of roots α+α=a−2b )
∵ Both equations have one root in common.
∴ Common root is −ab. It will satisfy equation.
∴d(a−b)2+2e(−b)a+f=0⇒db2−2aeb+a2f=0
⇒d(ac)−2aeb+a2f=0 (∵b2=ac)
⇒dc−2aeb+af=0
⇒acdc−ac2eb+acaf=0 (dividing by ac )
⇒ad−b22eb+cf=−⇒ad+cf=b2e
⇒ad,be,cf are in A.P.