Let $S_n$ be the sum to $n$-terms of an arithmetic progression $3, 7, 11, \ldots$, if $40 <…

Given, $\mathrm{AP}: 3,7,11,15,...$

$\Rightarrow a_n=3+\left(n-1\right)\left(4\right)$

$\Rightarrow a_n=4n-1$

$\Rightarrow S_n=\frac{n}{2}\left[6+4n-4\right]$

$\Rightarrow S_n=\frac{n\times 2\left(2n+1\right)}{2}$

$\Rightarrow S_n=2n^2+n$

Now, solving

$\Rightarrow \sum _{k=1}^n{S}_k=2\sum _{n=1}^n{n}^2+\sum _{n=1}^{n}n$

$\Rightarrow \sum _{k=1}^n{S}_k=\frac{n\left(n+1\right)\left(2n+1\right)}{3}+\frac{n\left(n+1\right)}{2}$

$\Rightarrow \sum _{k=1}^n{S}_k=\frac{n\left(n+1\right)}{2}\left[1+\frac{4n+2}{3}\right]$

$40 < \frac{6}{n(n+1)}\sum_{k=1}^{n}S_{k} < 42$ $\Rightarrow 40 < \frac{6}{n(n+1)} \times \frac{n(n+1)}{2} \left[\frac{5+4n}{3}\right] < 42$ $\Rightarrow 40 < 4n+5 < 42$ $\Rightarrow 35 < 4n < 37$ $\Rightarrow n = 9$