If ${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx}$, then :

$${I_n} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^n}xdx} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{n - 2}}x(\cos e{c^2}x - 1)dx} $$ <br><br>= $$\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx$$ <br><br>$$ = \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}$$<br><br>$= {1 \over {n - 1}} - {I_{n - 2}}$<br><br>$\Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}$<br><br>$\Rightarrow {I_2} + {I_4} = {1 \over 3}$<br><br>${I_3} + {I_5} = {1 \over 4}$<br><br>${I_4} + {I_6} = {1 \over 5}$<br><br>$\therefore$ ${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$ are in A.P.