Question

If $a_n={\int }_0^{\pi /2}\frac{{\sin }^{2}nx}{\sin x}dx$ , 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)

Since $a_n=\underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{2}nx}{sinx}dx$
$a_n-a_{n-1}=\underset{0}{\overset{\pi /2}{\int }}\frac{si{n}^{2}nx-si{n}^2\left(n-1\right)x}{sinx}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\frac{1-cos2nx}{2}\right)-\left(\frac{1-cos2\left(n-1\right)x}{2}\right)}{sinx}dx$
$=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\frac{cos2\left(n-1\right)x-cos2nx}{sinx}dx$
$=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\frac{2sin\left(2n-1\right)x-sinx}{sinx}dx$
$=\underset{0}{\overset{\pi /2}{\int }}sin\left(2n-1\right)xdx$
$={\left|-\frac{cos\left(2n-1\right)x}{2n-1}\right|}_0^{\pi /2}$
$=\frac{1}{2n-1}\left[-0+1\right]=\frac{1}{2n-1}$
$\therefore a_2-a_1=\frac{1}{3}$
$a_3-a_2=\frac{1}{5}$
$a_4-a_3=\frac{1}{7}.....$
$\therefore a_2-a_1,a_3-a_2,a_4-a_3,.....$ are in H.P.
[$\because 3,5,7,.....$ form an A.P.]