Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3,7,11, \ldots \ldots$.
If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then n equals
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