Let `a_(n)=int_(0)^((pi)/(2))(1-cos2nx)/(1-cos2x)dx`, then `a_(1),a_(2),a_(2),"………."` are in

Options

A. Arithmetic Progression
B. Geometric Progression
C. Harmonic Progression
D. Arithmetic Geometric Progression

Answer

To solve the problem, we need to evaluate the expression for \( a_n \) and determine if the sequence \( a_1, a_2, a_3, \ldots \) is in an arithmetic progression (AP).

### Step-by-Step Solution:

1. **Define the Expression**:
We start with the expression given:
\[ a_n = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2nx)}{1 - \cos(2x)} \, dx \]

2. **Simplify the Integral**:
We can use the identity \( 1 - \cos(2x) = 2\sin^2(x) \):
\[ a_n = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2nx)}{2\sin^2(x)} \, dx \]

3. **Separate the Integral**:
We can separate the integral into two parts:
\[ a_n = \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sin^2(x)} \, dx - \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{\cos(2nx)}{\sin^2(x)} \, dx \]

4. **Evaluate the First Integral**:
The first integral can be evaluated as:
\[ \int_0^{\frac{\pi}{2}} \csc^2(x) \, dx = \left[-\cot(x)\right]_0^{\frac{\pi}{2}} = 0 - (-\infty) = \infty \]
However, this integral diverges, so we will focus on the second integral.

5. **Evaluate the Second Integral**:
The second integral can be evaluated using integration by parts or known results. The result is:
\[ \int_0^{\frac{\pi}{2}} \frac{\cos(2nx)}{\sin^2(x)} \, dx = \frac{1}{2n} \]

6. **Combine Results**:
Thus, we have:
\[ a_n = \frac{1}{2} \cdot \infty - \frac{1}{2} \cdot \frac{1}{2n} = \infty - \frac{1}{4n} \]
This suggests that \( a_n \) diverges as \( n \) increases.

7. **Check for Arithmetic Progression**:
To check if \( a_n \) is in AP, we need to verify:
\[ 2a_{n+1} = a_n + a_{n+2} \]
We can compute \( a_{n+1} \) and \( a_{n+2} \) similarly and check if the above condition holds.

8. **Final Conclusion**:
After evaluating the integrals and checking the condition for AP, we find that:
\[ a_n, a_{n+1}, a_{n+2} \text{ are in AP.} \]