Let S _ n be the sum to n-terms of an arithmetic progression 3, 7,...

Given, $AP: 3, 7, 11, 15, . . .$

$⇒ {a}_{n} = 3 + \left(n - 1\right)\left(4\right)$

$⇒ {a}_{n} = 4n - 1$

$⇒ {S}_{n} = \dfrac{n}{2}\left[6 + 4n - 4\right]$

$⇒ {S}_{n} = \dfrac{n \times 2\left(2n + 1\right)}{2}$

$⇒ {S}_{n} = 2{n}^{2} + n$

Now, solving

$⇒ \mathop{∑}_{k = 1}^{n}{S}_{k} = 2\mathop{∑}_{n = 1}^{n}{n}^{2} + \mathop{∑}_{n = 1}^{n}n$

$⇒ \mathop{∑}_{k = 1}^{n}{S}_{k} = \dfrac{n\left(n + 1\right)\left(2n + 1\right)}{3} + \dfrac{n\left(n + 1\right)}{2}$

$⇒ \mathop{∑}_{k = 1}^{n}{S}_{k} = \dfrac{n\left(n + 1\right)}{2}\left[1 + \dfrac{4n + 2}{3}\right]$

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

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

$⇒ 40 < 4n + 5 < 42$

$⇒ 35 < 4n < 37$

$⇒ n = 9$