Let ` (1 + x^(2))^(2) (1 + x)^(n) = a_(0) + a_(1) x + a_(2) x^(2) + …` if
` a_(1),a_(2) " and " a_(3)` are in A.P , the value of n is

Options

A. 2
B. 3
C. 4
D. 7

Explanation

To solve the problem, we need to find the value of \( n \) such that the coefficients \( a_1, a_2, \) and \( a_3 \) from the expansion of \( (1 + x^2)^2 (1 + x)^n \) are in arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Expand the expression**:
We start with the expression \( (1 + x^2)^2 (1 + x)^n \).
First, we expand \( (1 + x^2)^2 \):
\[ (1 + x^2)^2 = 1 + 2x^2 + x^4 \]
Thus, the expression becomes:
\[ (1 + 2x^2 + x^4)(1 + x)^n \]

2. **Expand \( (1 + x)^n \)**:
Using the Binomial Theorem, we can expand \( (1 + x)^n \):
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]

3. **Combine the expansions**:
Now we multiply the two expansions:
\[ (1 + 2x^2 + x^4) \sum_{k=0}^{n} \binom{n}{k} x^k \]
This gives us:
\[ = \sum_{k=0}^{n} \binom{n}{k} x^k + 2 \sum_{k=0}^{n} \binom{n}{k} x^{k+2} + \sum_{k=0}^{n} \binom{n}{k} x^{k+4} \]

4. **Identify coefficients**:
From the combined expansion, we need to identify the coefficients \( a_1, a_2, \) and \( a_3 \):
- \( a_1 \) is the coefficient of \( x^1 \):
\[ a_1 = \binom{n}{1} = n \]
- \( a_2 \) is the coefficient of \( x^2 \):
\[ a_2 = \binom{n}{2} + 2\binom{n}{0} = \frac{n(n-1)}{2} + 2 \]
- \( a_3 \) is the coefficient of \( x^3 \):
\[ a_3 = \binom{n}{3} + 2\binom{n}{1} = \frac{n(n-1)(n-2)}{6} + 2n \]

5. **Set up the A.P. condition**:
For \( a_1, a_2, a_3 \) to be in A.P., we need:
\[ 2a_2 = a_1 + a_3 \]
Substituting the values we found:
\[ 2\left(\frac{n(n-1)}{2} + 2\right) = n + \left(\frac{n(n-1)(n-2)}{6} + 2n\right) \]

6. **Simplify the equation**:
Simplifying the left side:
\[ n(n-1) + 4 \]
Simplifying the right side:
\[ n + \frac{n(n-1)(n-2)}{6} + 2n = 3n + \frac{n(n-1)(n-2)}{6} \]
Setting both sides equal:
\[ n(n-1) + 4 = 3n + \frac{n(n-1)(n-2)}{6} \]

7. **Clear the fraction**:
Multiply through by 6 to eliminate the fraction:
\[ 6n(n-1) + 24 = 18n + n(n-1)(n-2) \]
Rearranging gives:
\[ n(n-1)(n-2) - 6n(n-1) + 18n - 24 = 0 \]

8. **Factor the polynomial**:
This leads to a cubic equation. After simplifying, we can factor or use the rational root theorem to find potential solutions. Testing \( n = 2, 3, 4 \) gives us valid solutions.

9. **Final values**:
The values of \( n \) that satisfy the conditions are \( n = 3 \) and \( n = 4 \) (since \( n = 2 \) does not allow for \( a_3 \)).

### Conclusion:
The values of \( n \) that satisfy the condition that \( a_1, a_2, a_3 \) are in A.P. are \( n = 3 \) and \( n = 4 \).