Given,

$b_n={\int }_0^{\frac{\pi }{2}}\frac{1+\cos 2nx}{\sin x}dx$

$b_{\mathrm{n}+1}-b_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^2\left(n+1\right)x-{\cos }^{2}nx}{\sin x}dx$

$={\int }_0^{\frac{\pi }{2}}\frac{-\sin \left(2n+1\right)x\sin x}{\sin x}dx$

$={\left(\frac{\cos (2n+1)x}{2n+1}\right)}_0^{\frac{\pi }{2}}=\frac{-1}{2n+1}$

So,$\frac{1}{b_3-b_2}=-5$

$\frac{1}{b_4-b_3}=-7$

$\frac{1}{b_5-b_4}=-9$

So,$\frac{1}{b_3-b_2},\frac{1}{b_4-b_3},\frac{1}{b_5-b_4}$ are in A.P. with $c.d=-2$