If an=∫0π/2 sin2⁡nxsin⁡xdx then a2−a1,a3−a2,a4−a3 are in

n>1,an−an−1=∫0π/2 sin2⁡nx−sin2⁡(n−1)xsin⁡xdx=∫0π/2 sin⁡(2n−1)xsin⁡xsin⁡xdx=∫0π/2 sin⁡(2n−1)xdx=−cos⁡(2n−1)x2n−10π/2=12n−1which is a Harmonic Progression.

If an=∫0π/2 sin2⁡nxsin⁡xdx then a2−a1,a3−a2,a4−a3 are in

n>1,an−an−1=∫0π/2 sin2⁡nx−sin2⁡(n−1)xsin⁡xdx=∫0π/2 sin⁡(2n−1)xsin⁡xsin⁡xdx=∫0π/2 sin⁡(2n−1)xdx=−cos⁡(2n−1)x2n−10π/2=12n−1which is a Harmonic Progression.

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$I f a_{n} = \int_{0}^{π / 2} \frac{\sin^{2} n x}{\sin x} d x$ then $a_{2} - a_{1}, a_{3} - a_{2},$
$a_{4} - a_{3}$ are in

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Detailed Solution

$n > 1, a_{n} - a_{n - 1} = \int_{0}^{π / 2} \frac{\sin^{2} n x - \sin^{2} (n - 1) x}{\sin x} d x$

$\begin{array}{l} = \int_{0}^{π / 2} \frac{\sin (2 n - 1) x \sin x}{\sin x} d x \\ = \int_{0}^{π / 2} \sin (2 n - 1) x d x = - {\frac{\cos (2 n - 1) x}{2 n - 1}|}_{0}^{π / 2} \\ = \frac{1}{2 n - 1} \end{array}$