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

<p>$$\begin{aligned} & \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\ &=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\ &=2 \mathrm{n}^2+\mathrm{n} \\ & \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K}^2+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K} \\ &=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ &=\mathrm{n}(\mathrm{n}+1)\left[\frac{2 \mathrm{n}+1}{3}+\frac{1}{2}\right] \\ &=\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}+5)}{6} \\ & \Rightarrow 40<\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{k=1}^n S_{\mathrm{k}}<42 \\ & 40<4 \mathrm{n}+5<42 \\ & 35<4 n<37 \\ & \mathrm{n}=9 \end{aligned}$$</p>