Let $\mathrm{a}_{\mathrm{n}}$ be the $\mathrm{n}^{\text {th }}$ term of the series $5+8+14+23+35+50+\ldots$ and $\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$. Then $\mathrm{S}_{30}-a_{40}$ is equal to :

Let $\mathrm{S}_n=5+8+14+23+\ldots .+a_n$ <br/><br/>and $\mathrm{S}_n=0+5+8+14+\ldots .+a_n$ <br/><br/>On subtracting, we get <br/><br/>$$ \begin{aligned} & 0=5+3+6 \ldots-a_n \\\\ & \Rightarrow a_n=5+3+6+9+\ldots(n-1) \text { terms } \\\\ & =5+\left[\frac{(n-1)}{2}(6+(n-2) 3)\right] \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =5+\left[\frac{(n-1)}{2}(6+3 n-6)\right] \\\\ & =5+\frac{(n-1)(3 n)}{2} \\\\ & =\frac{10+3 n^2-3 n}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { So, } a_{40}=\frac{3(40)^2-3(40)+10}{2} \\\\ & =\frac{4800-120+10}{2}=2345 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Now, } S_n=\sum_{k=1}^n a_k \\\\ & \Rightarrow S_{30}=\frac{3 \sum_{n=1}^{30} n^2-3 \sum_{n=1}^{30} n+10 \sum_{n=1}^{30} 1}{2} \\\\ & =\frac{3 \times(30)(30+1)(60+1)}{12}-\frac{3 \times 30 \times 31}{4} +\frac{10 \times 30}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & =\frac{28365-1395+300}{2}=\frac{27270}{2} \\\\ & =13635 \\\\ & \therefore S_{30}-a_{40}=13635-2345=11290 \end{aligned} $$