Since an=0∫π/2sinxsin2nxdx an−an−1=0∫π/2sinxsin2nx−sin2(n−1)xdx =0∫π/2sinx(21−cos2nx)−(21−cos2(n−1)x)dx =210∫π/2sinxcos2(n−1)x−cos2nxdx =210∫π/2sinx2sin(2n−1)x−sinxdx =0∫π/2sin(2n−1)xdx =∣∣−2n−1cos(2n−1)x∣∣0π/2 =2n−11[−0+1]=2n−11 ∴a2−a1=31 a3−a2=51 a4−a3=71..... ∴a2−a1,a3−a2,a4−a3,..... are in H.P.
[∵3,5,7,..... form an A.P.]