Let \( < a_n
> \)be a sequence such that \( a_1 + a_2 + \ldots + a_n = \dfrac{n^2 + 3n}{(n+1)(n+2)} \). If \( 28 \displaystyle\sum_{k=1}^{10} \dfrac{1}{a_k} = p_1 p_2 p_3 \ldots p_m \), where \( p_2, p_2, \ldots , p_m \) are the first \( m \) prime numbers, then \( m \) is equal to:
