The correct option is B A . P
In=Iπ/40tann−1xtan2xdx=Iπ/40tann−1xsec2x−12xdxIn=[tann+1xn +1]π40−In−2
∴In+In−2=1n−1
Putting n = 4 , 5 , 6 .... we get
I4+I2+I5+I3+I6+I4..... are respectivily 131415 Which are in H . P . are hence their reciprocals are in A . P