Let In=∫0π/4 tannxdx. Then I2+I4,I3+I5,I4+I6,I5+I7,…are in
A.P.
G.P.
H.P.
none of these
We have for r≥2,Ir+Ir+2=∫0π/4 tanrx1+tan2xdx =∫0π/4 tanrxsec2xdx =1r+1tanr+1x0π/4=1r+1Thus, the given sequence becomes, 13,14,15,16,…
This is clearly an H.P.