If `(1+x+x^(2))^(n)=a_(0)+a_(1)x+a_(2)x^(2)+….+a_(2n)x^(2n)` where `a_(0)`, `a(1)`, `a(2)` are unequal and in A.P., then `(1)/(a_(n))` is equal to :

Options

A. `(1)/((2n-1))`
B. `(1)/((2n+1))`
C. `(2)/(2n-1)`
D. `(1)/(n+2)`

Answer

To solve the problem, we need to analyze the expression \((1 + x + x^2)^n\) and find the coefficients \(a_0\), \(a_1\), and \(a_2\) such that they are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Rewrite the Expression**:
\[ (1 + x + x^2)^n \]
This can be expressed as:
\[ (1 + x)(1 + x^2)^n \]

2. **Expand Using the Binomial Theorem**:
We can expand \((1 + x + x^2)^n\) using the multinomial theorem, but for our purpose, we will focus on the coefficients:
- \(a_0\) is the constant term.
- \(a_1\) is the coefficient of \(x\).
- \(a_2\) is the coefficient of \(x^2\).

3. **Identify the Coefficients**:
- The coefficient \(a_0\) (the constant term) is simply \(1\) (when \(x = 0\)).
- The coefficient \(a_1\) (the coefficient of \(x\)) can be found using:
\[ a_1 = \binom{n}{1} = n \]
- The coefficient \(a_2\) (the coefficient of \(x^2\)) can be calculated as:
\[ a_2 = \binom{n}{2} + \binom{n}{1} = \frac{n(n-1)}{2} + n = \frac{n(n+1)}{2} \]

4. **Set Up the A.P. Condition**:
Since \(a_0\), \(a_1\), and \(a_2\) are in A.P., we have:
\[ 2a_1 = a_0 + a_2 \]
Substituting the values:
\[ 2n = 1 + \frac{n(n + 1)}{2} \]

5. **Solve the Equation**:
Multiply through by \(2\) to eliminate the fraction:
\[ 4n = 2 + n(n + 1) \]
Rearranging gives:
\[ n^2 - 3n + 2 = 0 \]
Factoring:
\[ (n - 1)(n - 2) = 0 \]
Thus, \(n = 1\) or \(n = 2\).

6. **Determine Valid \(n\)**:
Since \(a_0\), \(a_1\), and \(a_2\) are unequal, we discard \(n = 1\). Therefore, \(n = 2\).

7. **Find \(a_n\)**:
Now we need to find \(a_2\):
\[ a_2 = \frac{2(2 + 1)}{2} = 3 \]

8. **Calculate \(\frac{1}{a_n}\)**:
Finally, we compute:
\[ \frac{1}{a_2} = \frac{1}{3} \]

### Final Answer:
\[ \frac{1}{a_n} = \frac{1}{3} \]