Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3, 7, 11, \ldots$, if $40 < \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k < 42$, then $n$ equals $\underline{\hspace{1cm}}$.
Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3, 7, 11, \ldots$, if $40 < \frac{6}{n(n+1)}\sum_{k=1}^{n} S_k < 42$, then $n$ equals $\underline{\hspace{1cm}}$.