Let `n in N`. If `(1+x)^(n)=a_(0)+a_(1)x+a_(2)x^(2)+…….+a_(n)x^(n)` and `a_(n)-3,a_(n-2), a_(n-1)` are in AP, then :

Options

A. `a_(1),a_(2),a_(3)` are in AP
B. `a_(1),a_(2),a_(3)` are in HP
C. `n=17`
D. `n=14`

Answer

To solve the problem, we need to analyze the given information and apply the properties of the coefficients of the binomial expansion.

### Step-by-Step Solution:

1. **Understanding the Binomial Expansion**:
The binomial theorem states that:
\[ (1 + x)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n \]
where \( a_k = \binom{n}{k} \).

2. **Identifying the Coefficients**:
From the binomial expansion, we have:
\[ a_{n-3} = \binom{n}{n-3}, \quad a_{n-2} = \binom{n}{n-2}, \quad a_{n-1} = \binom{n}{n-1} \]

3. **Condition for Arithmetic Progression (AP)**:
The coefficients \( a_{n-3}, a_{n-2}, a_{n-1} \) are in AP if:
\[ 2a_{n-2} = a_{n-3} + a_{n-1} \]

4. **Substituting the Coefficients**:
Substitute the binomial coefficients into the AP condition:
\[ 2 \binom{n}{n-2} = \binom{n}{n-3} + \binom{n}{n-1} \]
This simplifies to:
\[ 2 \binom{n}{2} = \binom{n}{3} + \binom{n}{1} \]

5. **Using the Binomial Coefficient Formulas**:
Recall that:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Therefore:
\[ \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{3} = \frac{n(n-1)(n-2)}{6}, \quad \binom{n}{1} = n \]

6. **Setting Up the Equation**:
Substitute these into the equation:
\[ 2 \cdot \frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6} + n \]
This simplifies to:
\[ n(n-1) = \frac{n(n-1)(n-2)}{6} + n \]

7. **Clearing the Fraction**:
Multiply through by 6 to eliminate the fraction:
\[ 6n(n-1) = n(n-1)(n-2) + 6n \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 6n^2 - 6n = n^3 - 3n^2 + 2n \]
Simplifying further:
\[ n^3 - 9n^2 + 8n = 0 \]

9. **Factoring the Polynomial**:
Factor out \( n \):
\[ n(n^2 - 9n + 8) = 0 \]
The quadratic can be factored as:
\[ n(n-1)(n-8) = 0 \]

10. **Finding the Natural Number Solutions**:
The solutions are \( n = 0, n = 1, n = 8 \). Since \( n \in \mathbb{N} \), we take \( n = 8 \).

### Conclusion:
The value of \( n \) such that \( a_{n-3}, a_{n-2}, a_{n-1} \) are in AP is \( n = 8 \).