Answer is AP .

draw triangle abc

draw an altitude on ac from b such that it intersects ac at s and intersects a line l parallel to ac going thru the triangle at H

(this is the orthocentre)

now draw the perpendicular bisector of ac intersecting ac at r and line l at L

(this is the circumcentre)

similarly draw the median intersecting ac at r and line l at 

G(this is the centroid...the centroid will also lie on line l as it divides join of H and L (circumcentre) i ratio 2:1) 

now consider triangles bgh and glr

gl:hg = 2:1 (centroid,circumcentre and orthocentre prop.) 

bg:gr = 2:1 (centroid property that it divides median into 2:1 ratio)

so the two triangles are similar

so we can say

bh:lr = 2:1 = bh:hs = 2RcosB/2RcosAcosC(this is the property for orthocentre and altitude)

so we get 2cosAcosC = cosB

 2cosAcosC = -cos(A+C)

so we get on simplification

tanAtanC = 3 .....(1)

now tan(A+C) = -tan(B)

(tanA + tanC)/(1-tanAtanC) = -tanB

use (1) and get

tanA +tanC = 2tanB

so AP