Question

Let $a_n={\int }_0^{\pi /2}\frac{{\sin }^{2}nx}{\sin x}dx$. Then $a_2-a_1,a_3-a_2,a_4-a_3$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of these

Answer 1

Answer: (C)

$H.P.$

$a_n-a_{n-1}={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}nx}{\sin x}dx-{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2(n-1)x}{\sin x}dx$

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

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

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

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

$={\int }_0^{\frac{\pi }{2}}\sin (2n-1)xdx$

$={[-\frac{\cos (2n-1)x}{2n-1}]}_0^{\frac{\pi }{2}}=-\frac{1}{2n-1}[0-1]$

$=\frac{1}{2n-1}$

$\therefore a_2-a_1=\frac{1}{3},a_3-a_2=\frac{1}{5},a_4-a_3=\frac{1}{7},...$ which are in $H.P.$

$[\because 3,5,7,...$ are in $A.P.]$