Mathematics - Sequence & Series Question with Solution | TestHub

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

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

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

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

Now, using $40<\frac{6}{n\left(n+1\right)}\sum _{k=1}^n{S}_k<42$ we get,

$\Rightarrow 40<\frac{6}{n\left(n+1\right)}\times \frac{n\left(n+1\right)}{2}\left[\frac{5+4n}{3}\right]<42$

$\Rightarrow 40<4n+5<42$

$\Rightarrow 35<4n<37$

$\Rightarrow n=9$