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 start with the expression given:

\[ (1 + x + x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n} \]

where \( a_0, a_1, a_2 \) are unequal and in arithmetic progression (A.P.).

### Step 1: Rewrite the expression
We can rewrite \( 1 + x + x^2 \) as follows:

\[ 1 + x + x^2 = 1 + x(1 + x) \]

This allows us to express \( (1 + x + x^2)^n \) in a more manageable form.

### Step 2: Use the Binomial Theorem
We can use the Binomial Theorem to expand \( (1 + x + x^2)^n \). We can treat \( 1 + x + x^2 \) as a single term and expand it:

\[ (1 + x + x^2)^n = \sum_{k=0}^{n} \binom{n}{k} (x + x^2)^k \]

### Step 3: Expand \( (x + x^2)^k \)
Now we need to expand \( (x + x^2)^k \):

\[ (x + x^2)^k = \sum_{j=0}^{k} \binom{k}{j} x^j (x^2)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} x^{j + 2(k-j)} = \sum_{j=0}^{k} \binom{k}{j} x^{2k - j} \]

### Step 4: Collect coefficients
From the above expansion, we can collect the coefficients of \( x^0, x^1, x^2, \ldots, x^{2n} \). The coefficients \( a_k \) can be expressed in terms of binomial coefficients.

### Step 5: Set up the A.P. condition
Given that \( a_0, a_1, a_2 \) are in A.P., we have:

\[ 2a_1 = a_0 + a_2 \]

From our expansion, we know:

- \( a_0 = 1 \) (the coefficient of \( x^0 \))
- \( a_1 = \binom{n}{1} \)
- \( a_2 = \binom{n}{1} + \binom{n}{2} \)

Substituting these into the A.P. condition gives:

\[ 2 \binom{n}{1} = 1 + \left( \binom{n}{1} + \binom{n}{2} \right) \]

### Step 6: Solve for \( n \)
This simplifies to:

\[ 2n = 1 + n + \frac{n(n-1)}{2} \]

Multiplying through by 2 to eliminate the fraction:

\[ 4n = 2 + 2n + n(n-1) \]

Rearranging gives:

\[ n^2 - 2n - 2 = 0 \]

### Step 7: Use the quadratic formula
Using the quadratic formula:

\[ n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3} \]

### Step 8: Determine valid \( n \)
Since \( n \) must be a positive integer, we take \( n = 2 \).

### Step 9: Find \( a_n \)
Now, we need to find \( a_n \):

\[ a_2 = \binom{2}{1} + \binom{2}{2} = 2 + 1 = 3 \]

### Step 10: Calculate \( \frac{1}{a_n} \)
Finally, we find:

\[ \frac{1}{a_n} = \frac{1}{3} \]

### Conclusion
Thus, the answer is:

\[ \frac{1}{a_n} = \frac{1}{3} \]