Q. If $b _{ n }=\int\limits_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} nx }{\sin x } dx , n \in N$, then

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A

$b_{3}-b_{2}, b_{4}-b_{3}, b_{5}-b_{4}$ are in an A.P. with common difference $-2$

B

$\frac{1}{ b _{3}- b _{2}}, \frac{1}{ b _{4}- b _{3}}, \frac{1}{ b _{5}- b _{4}}$ are in an A.P. with common difference 2

C

$b_{3}-b_{2}, b_{4}-b_{3}, b_{5}-b_{4}$ are in a G.P.

D

$\frac{1}{ b _{3}- b _{2}}, \frac{1}{ b _{4}- b _{3}}, \frac{1}{ b _{5}- b _{4}}$ are in an A.P. with common difference $-2$

Solution:

$b _{ n }=\int\limits_{0}^{\pi / 2} \frac{1+\cos 2 nx }{\sin x } dx$
$b _{ n +1}- b _{ n }=\int\limits_{0}^{\pi / 2} \frac{\cos ^{2}( n +1) x -\cos ^{2} nx }{\sin x } dx$
$=\int\limits_{0}^{\pi / 2} \frac{-\sin (2 n +1) x \sin x }{\sin x } dx$
$=\left(\frac{\cos (2 n +1) x }{2 n +1}\right)_{0}^{\pi / 2}=\frac{-1}{2 n +1}$
$\frac{1}{ b _{3}- b _{2}}, \frac{1}{ b _{4}- b _{3}}, \frac{1}{ b _{5}- b _{4}}$ are in A.P. with c.d. $=-2$

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