If b _ n = ∫ _ 0 ^ (π)/(2) ^ 2 nx x dx, n ∈ ℕ , then

Given,

${b}_{n} = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{1 + \cos 2nx}{\sin x}dx$

${b}_{n + 1} - {b}_{n} = {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\cos^{2}\left(n + 1\right)x - \cos^{2}nx}{\sin x}dx$

$= {∫}_{0}^{\dfrac{\pi}{2}}\dfrac{- \sin \left(2n + 1\right)xsinx}{\sin x}dx$

$= {\left(\dfrac{\cos ( 2n + 1 ) x}{2n + 1}\right)}_{0}^{\dfrac{\pi}{2}} = \dfrac{- 1}{2n + 1}$

So,$\dfrac{1}{{b}_{3} - {b}_{2}} = - 5$

$\dfrac{1}{{b}_{4} - {b}_{3}} = - 7$

$\dfrac{1}{{b}_{5} - {b}_{4}} = - 9$

So,$\dfrac{1}{{b}_{3} - {b}_{2}}, \dfrac{1}{{b}_{4} - {b}_{3}}, \dfrac{1}{{b}_{5} - {b}_{4}}$ are in A.P. with $c . d = - 2$