$a_{n+2}+a_n-2a_{n+1}=$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-\cos 2\left(n+2\right)x}{1-\cos 2x}dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-\cos 2nx}{1-\cos 2x}dx-2\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-\cos 2\left(n+1\right)x}{1-\cos 2x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\left(\frac{1-\cos \left(2n+4\right)x+1-\cos 2nx-2+2\cos \left(2n+2\right)x}{1-\cos 2x}\right)dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{2\cos \left(2n+2\right)x-\left(\cos \left(2n+4\right)x+\cos 2nx\right)}{1-\cos 2x}dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{2\cos \left(2n+2\right)x-2\cos \left(2n+2\right)x\cos 2x}{1-\cos 2x}dx$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{2\cos \left(2n+2\right)x\cancel{\left(1-\cos 2x\right)}}{\cancel{\left(1-\cos 2x\right)}}dx$
${\left.=\frac{\cancel{2}\sin \left(2n+2\right)x}{\cancel{2}\left(n+1\right)}\right|}_0^{\frac{\pi }{2}}=\frac{1}{n+1}\left(\sin \left(n+1\right)\pi -\sin 0\right)=0$
$\Rightarrow a_{n+2}+a_n=2a_{n+1}$
$\Rightarrow$ They are in $\mathrm{A}.\mathrm{P}.$