If I_n=∫ _ π 4^ π 2 cot^nxdx, then

$${I}_{n}={\int }_{\pi /4}^{\pi /2}{\mathrm{cot}}^{n}xdx={\int }_{\pi /4}^{\pi /2}{\mathrm{cot}}^{n-2}x({cosec}^{2}x-1)dx$$In​=∫π/4π/2​cotnxdx=∫π/4π/2​cotn−2x(cosec2x−1)dx

$${=-\frac{{\mathrm{cot}}^{n-1}x}{n-1}]}_{\pi /4}^{\pi /2}-{I}_{n-2}$$=−n−1cotn−1x​]π/4π/2​−In−2​

$$=\frac{1}{n-1}-{I}_{n-2}$$=n−11​−In−2​

$$\Rightarrow {I}_{n}+{I}_{n-2}=\frac{1}{n-1}$$⇒In​+In−2​=n−11​

$$\Rightarrow {I}_{2}+{I}_{4}=\frac{1}{3}$$⇒I2​+I4​=31​

$${I}_{3}+{I}_{5}=\frac{1}{4}$$I3​+I5​=41​

$${I}_{4}+{I}_{6}=\frac{1}{5}$$I4​+I6​=51​

$\therefore \frac{1}{{I}_{2}+{I}_{4}},\frac{1}{{I}_{3}+{I}_{5}},\frac{1}{{I}_{4}+{I}_{6}}$∴I2​+I4​1​,I3​+I5​1​,I4​+I6​1​ are in $A.P.$A.P.