Question

If $a_n=\displaystyle \int_{0} ^{\pi/2}\frac{\sin^2nx}{\sin\,x}dx$ , 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 $a_{n}= \int\limits_{0}^{\pi /2} \frac{sin^{2}\,nx}{sin\,x}dx$
$a_{n}-a_{n-1}=\int\limits_{0}^{\pi /2} \frac{sin^{2}\,nx-sin^{2}\left(n-1\right)x}{sin\,x}dx$
$=\int\limits_{0}^{\pi /2} \frac{\left(\frac{1-cos\,2\,nx}{2}\right)-\left(\frac{1-cos\,2\left(n-1\right)x}{2}\right)}{sin\, x} dx$
$=\frac{1}{2} \int\limits_{0}^{\pi/ 2} \frac{cos\,2\left(n-1\right)x-cos\,2\,nx}{sin\,x} dx$
$=\frac{1}{2} \int\limits_{0}^{\pi /2} \frac{2 sin\left(2n-1\right)x-sin\,x}{sin\,x}dx$
$=\int\limits_{0}^{\pi /2} sin \left(2n-1\right)x\, dx$
$=\left|-\frac{cos\left(2n-1\right)x}{2n-1}\right|_{0}^{\pi /2}$
$=\frac{1}{2n-1}\left[-0+1\right]=\frac{1}{2n-1}$
$\therefore a_{2}-a_{1}=\frac{1}{3}$
$a_{3}-a_{2}=\frac{1}{5}$
$a_{4}-a_{3}=\frac{1}{7} .....$
$\therefore a_{2}-a_{1}, a_{3}-a_{2}, a_{4}-a_{3}, .....$ are in H.P.
[$\because3,5,7, .....$ form an A.P.]