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. $noneofthese$

Answer 1

The correct option is C $H.P.$

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

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

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

$={\int }_0^{\frac{\pi }{2}}\frac{\sin [nx+(n-1)x]\sin [nx-(n-1\left)x\right]}{\sin x}dx$ ....... $[\because {\cos }^{2}b-{\cos }^{2}a=\sin (a+b\left)\sin \right(a-b\left)\right]$

$={\int }_0^{\frac{\pi }{2}}\frac{\sin [2nx-x]\sin x}{\sin x}dx$

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

$2n-1$ is an AP then $\frac{1}{2n-1}$ is in HP