Let a n be n th term of the series 5 + 8 + 14 + 23 + 35 + 50 + . . . . . . . and S n = ∑ k = 1 n a k .…

The given series is $a_n$, then we can write as

$$\begin{gathered} a_n=5+8+14+23+............+a_n \\ {a}_n= 5+8+14+23+......+a_{n-1}+a_n \end{gathered}$$

Subtracting above equations, we get

$0=5+3+6+9+....+\left(a_n-a_{n-1}\right)-a_n$

$\Rightarrow a_n=5+\left[\frac{\left(n-1\right)}{2}\left(2\times 3+\left(n-1-1\right)3\right)\right]$

$\Rightarrow a_n=5+\frac{\left(n-1\right)}{2}\left(3n\right)$

$\Rightarrow a_n=\frac{1}{2}\left(3n^2-3n+10\right)$

So,

$\Rightarrow a_{40}=\frac{1}{2}\left(3\times {40}^2-3\times 40+10\right)$

$\Rightarrow a_{40}=\frac{1}{2}\left(4800-120+10\right)$

$\Rightarrow a_{40}=2345$

$\Rightarrow S_{30}=\sum _{n=1}^{30}{a}_n$

$\Rightarrow S_{30}=\frac{1}{2}\left(3\sum _{n=1}^{30}\left(n^2\right)-3\sum _{n=1}^{30}\left(n\right)+\sum _{n=1}^{30}10\right)$

$=\frac{1}{2}\left(\frac{3\left(30\right)\left(31\right)\left(61\right)}{6}-\frac{3\left(30\right)\left(31\right)}{2}+10\times 30\right)$

$\Rightarrow S_{30}=13635$

$\Rightarrow S_{30}-a_{40}=13635-2345=11290$

Therefore, the required value is $11290$.