If $\mathrm{S}_{n}=4+11+21+34+50+\ldots$ to $n$ terms, then $\frac{1}{60}\left(\mathrm{~S}_{29}-\mathrm{S}_{9}\right)$ is equal to :

Given that <br/><br/>$$ \begin{aligned} & \mathrm{S}_n=4+11+21+24+50+\ldots+\mathrm{T}_n \\\\ & \mathrm{~S}_n=~~~~~~~~4+11+21+34++\mathrm{T}_{n-1}+\mathrm{T}_n \\ & -\quad-\quad-\quad-\quad-\quad-\quad- \\ & \hline 0=4+7+10+13+16+\ldots\left(\mathrm{T}_n-\mathrm{T}_{n-2}\right)-\mathrm{T}_n \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \mathrm{T}_n=4+7+10+13+16+\ldots . \text { to } n \text { terms } \\\\ & \Rightarrow \mathrm{T}_n=\frac{n}{2}[2 \times 4+(n-1) 3] \\\\ & \mathrm{T}_n=\frac{3}{2} n^2+\frac{5}{2} n \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { So } \mathrm{S}_n=\Sigma \mathrm{T}_n=\frac{3}{2} \Sigma n^2+\frac{5}{2} \Sigma n \\\\ & \Rightarrow \mathrm{S}_n=\frac{3}{2} \times \frac{n(n+1)(2 n+1)}{6}+\frac{5}{2} \times \frac{n(n+1)}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow S_n=\frac{n(n+1)}{4}(2 n+1+5)=\frac{n(n+1)(n+3)}{2} \\\\ & \text { Hence, } \frac{1}{60}\left(S_{29}-S_9\right)=\frac{1}{60} \times \frac{1}{2}(29 \times 30 \times 32-9 \times 10 \times 12) \\\\ & =223 \end{aligned} $$