Find $k \in \mathbb{N}$ such that

[b]a.)[/b] For any $n \in \mathbb{N}$, there does not exist $j \in \mathbb{Z}$ which satisfies the conditions $0 \leq j \leq n - k + 1$ and $\left(
\begin{array}{c}
n\\
j\end{array} \right), \left( \begin{array}{c}
n\\
j + 1\end{array} \right), \ldots, \left( \begin{array}{c}
n\\
j + k - 1\end{array} \right)$ forms an arithmetic progression.

[b]b.)[/b] There exists $n \in \mathbb{N}$ such that there exists $j$ which satisfies $0 \leq j \leq n - k + 2$, and $\left(
\begin{array}{c}
n\\
j\end{array} \right), \left( \begin{array}{c}
n\\
j + 1\end{array} \right), \ldots , \left( \begin{array}{c}
n\\
j + k - 2\end{array} \right)$ forms an arithmetic progression.

Find all $n$ which satisfies part [b]b.)[/b]
### Part (a)
We need to find \( k \in \mathbb{N} \) such that for any \( n \in \mathbb{N} \), there does not exist \( j \in \mathbb{Z} \) which satisfies the conditions \( 0 \leq j \leq n - k + 1 \) and \( \binom{n}{j}, \binom{n}{j+1}, \ldots, \binom{n}{j+k-1} \) forms an arithmetic progression.

1. **Case \( k = 1 \)**:
   - If \( k = 1 \), then the sequence consists of a single binomial coefficient \( \binom{n}{j} \), which trivially forms an arithmetic progression.
   - Therefore, \( k = 1 \) does not satisfy the condition.

2. **Case \( k = 2 \)**:
   - If \( k = 2 \), then the sequence consists of two binomial coefficients \( \binom{n}{j} \) and \( \binom{n}{j+1} \).
   - Any two numbers form an arithmetic progression.
   - Therefore, \( k = 2 \) does not satisfy the condition.

3. **Case \( k \geq 3 \)**:
   - If \( k \geq 3 \), consider the sequence \( \binom{n}{j}, \binom{n}{j+1}, \binom{n}{j+2} \).
   - For these to form an arithmetic progression, the following must hold:
     \[
     \binom{n}{j} + \binom{n}{j+2} = 2 \binom{n}{j+1}
     \]
   - Using the properties of binomial coefficients, we have:
     \[
     \binom{n}{j} = \frac{n!}{j!(n-j)!}, \quad \binom{n}{j+1} = \frac{n!}{(j+1)!(n-j-1)!}, \quad \binom{n}{j+2} = \frac{n!}{(j+2)!(n-j-2)!}
     \]
   - Substituting these into the equation, we get:
     \[
     \frac{n!}{j!(n-j)!} + \frac{n!}{(j+2)!(n-j-2)!} = 2 \cdot \frac{n!}{(j+1)!(n-j-1)!}
     \]
   - Simplifying, we obtain:
     \[
     \frac{1}{j!(n-j)!} + \frac{1}{(j+2)!(n-j-2)!} = 2 \cdot \frac{1}{(j+1)!(n-j-1)!}
     \]
   - Further simplification leads to:
     \[
     (j+2)(j+1) + (n-j)(n-j-1) = 2(j+2)(n-j)
     \]
   - Expanding and simplifying:
     \[
     j^2 + 3j + 2 + n^2 - 2nj + j^2 - n + j = 2nj + 4n - 2j^2 - 4j
     \]
     \[
     4j^2 + (4n - 8)j + (n^2 - 5n + 2) = 0
     \]
   - Solving this quadratic equation for \( j \):
     \[
     j = \frac{n - 2 \pm \sqrt{n + 2}}{2}
     \]
   - For \( j \) to be an integer, \( n + 2 \) must be a perfect square. Let \( n + 2 = t^2 \) for some integer \( t \geq 2 \), then \( n = t^2 - 2 \).
   - Therefore, for any \( n \) and \( k \geq 3 \), there does not exist \( j \) such that the binomial coefficients form an arithmetic progression.

### Part (b)
We need to find \( k \in \mathbb{N} \) such that there exists \( n \in \mathbb{N} \) and \( j \) which satisfies \( 0 \leq j \leq n - k + 2 \) and \( \binom{n}{j}, \binom{n}{j+1}, \ldots, \binom{n}{j+k-2} \) forms an arithmetic progression.

1. **Case \( k = 1, 2, 3 \)**:
   - For \( k = 1 \), the sequence consists of a single binomial coefficient, which trivially forms an arithmetic progression.
   - For \( k = 2 \), any two binomial coefficients form an arithmetic progression.
   - For \( k = 3 \), we have already shown that for any \( n \), there exists \( j \) such that \( \binom{n}{j}, \binom{n}{j+1}, \binom{n}{j+2} \) can form an arithmetic progression.

2. **Case \( k \geq 4 \)**:
   - Similar to part (a), if \( k \geq 4 \), the sequence \( \binom{n}{j}, \binom{n}{j+1}, \ldots, \binom{n}{j+k-2} \) forms an arithmetic progression if \( n \) is of the form \( t^2 - 2 \) for some integer \( t \geq 2 \).

Conclusion:
- For \( k = 1, 2, 3 \), every \( n \) satisfies the condition.
- For \( k \geq 4 \), \( n \) must be of the form \( t^2 - 2 \) for some integer \( t \geq 2 \).

The final answer is \( \boxed{ n } \) must be of the form \( t^2 - 2 \) for some integer \( t \geq 2 \) when \( k \geq 4 \). For \( k = 1, 2, 3 \), every \( n \) satisfies the condition.