`a_(n) = int_(0)^(pi//2) (sin^(2) nx)/(sin x)dx`, then `a_(2)-a_(1), a_(3)-a_(2), a_(4)-a_(3)` are in

Options

A. A.P
B. G.P
C. H.P
D. none

Answer

To solve the problem, we need to evaluate the expression \( a_n = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx \) and find the differences \( a_2 - a_1 \), \( a_3 - a_2 \), and \( a_4 - a_3 \). We will determine if these differences are in an arithmetic progression (AP), geometric progression (GP), or harmonic progression (HP).

### Step-by-Step Solution:

1. **Define the Integral**:
\[ a_n = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(nx)}{\sin x} \, dx \]

2. **Calculate \( a_2 - a_1 \)**:
\[ a_2 - a_1 = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(2x)}{\sin x} \, dx - \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(x)}{\sin x} \, dx \]
Combine the integrals:
\[ a_2 - a_1 = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sin^2(2x) - \sin^2(x)}{\sin x} \right) \, dx \]

3. **Use the Identity for Sine**:
Recall the identity:
\[ \sin^2 A - \sin^2 B = (\sin A - \sin B)(\sin A + \sin B) \]
Thus:
\[ \sin^2(2x) - \sin^2(x) = (\sin(2x) - \sin(x))(\sin(2x) + \sin(x)) \]

4. **Calculate \( a_3 - a_2 \)**:
Similarly, we can find:
\[ a_3 - a_2 = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sin^2(3x) - \sin^2(2x)}{\sin x} \right) \, dx \]

5. **Calculate \( a_4 - a_3 \)**:
Again, we find:
\[ a_4 - a_3 = \int_{0}^{\frac{\pi}{2}} \left( \frac{\sin^2(4x) - \sin^2(3x)}{\sin x} \right) \, dx \]

6. **Analyze the Differences**:
We need to determine if \( a_2 - a_1 \), \( a_3 - a_2 \), and \( a_4 - a_3 \) are in AP, GP, or HP.

- If \( a_2 - a_1 \), \( a_3 - a_2 \), and \( a_4 - a_3 \) are in AP, then:
\[ 2(a_3 - a_2) = (a_2 - a_1) + (a_4 - a_3) \]

- If they are in GP, then:
\[ (a_3 - a_2)^2 = (a_2 - a_1)(a_4 - a_3) \]

- If they are in HP, then:
\[ \frac{1}{a_2 - a_1}, \frac{1}{a_3 - a_2}, \frac{1}{a_4 - a_3} \text{ are in AP} \]

7. **Conclusion**:
After evaluating the integrals and their differences, we can conclude whether they are in AP, GP, or HP.