If `I_(n)=int_(0)^(pi//4) tan^(n)x dx`, then
`(1)/(I_(2)+I_(4)),(1)/(I_(3)+I_(5)),(1)/(I_(4)+I_(6)),...` from\

Options

A. am A.P.
B. a G.P.
C. a H.P.
D. none of these

Explanation

To solve the problem, we need to evaluate the integrals \( I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \) and analyze the series formed by \( \frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \ldots \).

### Step-by-step Solution:

1. **Define the Integrals**:
We start with the integral defined as:
\[ I_n = \int_0^{\frac{\pi}{4}} \tan^n x \, dx \]

2. **Relate \( I_n \) and \( I_{n+2} \)**:
We can express \( I_n + I_{n+2} \):
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \, dx + \int_0^{\frac{\pi}{4}} \tan^{n+2} x \, dx \]
This can be combined into a single integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x (1 + \tan^2 x) \, dx \]

3. **Use the Identity**:
We know that \( 1 + \tan^2 x = \sec^2 x \). Thus, we can rewrite the integral:
\[ I_n + I_{n+2} = \int_0^{\frac{\pi}{4}} \tan^n x \sec^2 x \, dx \]

4. **Substitution**:
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \) and when \( x = 0 \), \( t = 0 \) and when \( x = \frac{\pi}{4} \), \( t = 1 \). The integral becomes:
\[ I_n + I_{n+2} = \int_0^1 t^n \, dt \]

5. **Evaluate the Integral**:
The integral \( \int_0^1 t^n \, dt \) can be evaluated as:
\[ \int_0^1 t^n \, dt = \frac{t^{n+1}}{n+1} \bigg|_0^1 = \frac{1}{n+1} \]
Therefore, we have:
\[ I_n + I_{n+2} = \frac{1}{n+1} \]

6. **Reciprocal Relation**:
Taking the reciprocal gives:
\[ \frac{1}{I_n + I_{n+2}} = n + 1 \]

7. **Calculate Specific Cases**:
- For \( n = 2 \):
\[ \frac{1}{I_2 + I_4} = 3 \]
- For \( n = 3 \):
\[ \frac{1}{I_3 + I_5} = 4 \]
- For \( n = 4 \):
\[ \frac{1}{I_4 + I_6} = 5 \]

8. **Identify the Sequence**:
The values obtained are:
\[ 3, 4, 5, \ldots \]
This sequence is an arithmetic progression (AP) with a common difference of 1.

### Conclusion:
Thus, the series \( \frac{1}{I_2 + I_4}, \frac{1}{I_3 + I_5}, \frac{1}{I_4 + I_6}, \ldots \) forms an arithmetic progression (AP).