1. Given integral

We have

$$I_n=\int_{\pi/4}^{\pi/2} \cot^n x\,dx.$$

We need to study the quantities:

$$I_2+I_4,\quad I_3+I_5,\quad I_4+I_6.$$

---

2. Transform the integral

Let

$$t=\cot x.$$

Then

$$dt=-\csc^2 x\,dx=-(1+\cot^2 x)dx=-(1+t^2)dx,$$

so

$$dx=-\frac{dt}{1+t^2}.$$

Also, when $x=\pi/4$, $t=1$, and when $x=\pi/2$, $t=0$. Hence

$$I_n=\int_{1}^{0} t^n\left(-\frac{dt}{1+t^2}\right)=\int_0^1 \frac{t^n}{1+t^2}\,dt.$$

So,

$$I_n=\int_0^1 \frac{t^n}{1+t^2}\,dt.$$

---

3. Compute the required sums

(i) $I_2+I_4$

$$I_2+I_4=\int_0^1 \frac{t^2+t^4}{1+t^2}\,dt =\int_0^1 t^2\,dt =\left[\frac{t^3}{3}\right]_0^1 =\frac13.$$

(ii) $I_3+I_5$

$$I_3+I_5=\int_0^1 \frac{t^3+t^5}{1+t^2}\,dt =\int_0^1 t^3\,dt =\left[\frac{t^4}{4}\right]_0^1 =\frac14.$$

(iii) $I_4+I_6$

$$I_4+I_6=\int_0^1 \frac{t^4+t^6}{1+t^2}\,dt =\int_0^1 t^4\,dt =\left[\frac{t^5}{5}\right]_0^1 =\frac15.$$

Thus the three numbers are

$$\frac13,\quad \frac14,\quad \frac15.$$

---

4. Check each option

Option A

The reciprocals are

$$\frac{1}{I_2+I_4}=3,\quad \frac{1}{I_3+I_5}=4,\quad \frac{1}{I_4+I_6}=5.$$

These are in A.P. since

$$4-3=1,\qquad 5-4=1.$$

So A is true.

Option B

The numbers themselves are

$$\frac13,\ \frac14,\ \frac15.$$

For A.P., middle term should be average of the other two:

$$\frac14 \stackrel{?}{=} \frac{\frac13+\frac15}{2}=\frac{8}{30}=\frac{4}{15}.$$

But

$$\frac14\ne \frac{4}{15}.$$

So B is false.

Option C

Check if reciprocals $3,4,5$ are in G.P. For G.P.,

$$4^2 \stackrel{?}{=} 3\cdot 5.$$

But

$$16\ne 15.$$

So C is false.

Option D

Check whether

$$I_2+I_4,\quad (I_3+I_5)^2,\quad I_4+I_6$$

are in G.P. These are

$$\frac13,\quad \left(\frac14\right)^2=\frac1{16},\quad \frac15.$$

For G.P.,

\left(\frac1{16}\right)^2 \stackrel{?}{=} \frac13\cdot \frac15=rac1{15}.

But

$$\frac1{256}\ne \frac1{15}.$$

So D is false.

---

5. Final answer

Only Option A is correct.