Given that, $a_{1},a_{2},a_{3}, \ldots, a_{3n} $ in AP.
And $a_{1} = 3, a_{2} = 7$
Common difference $(d) = a_{2} – a_{1} = 7-3 = 4$
The sum of $n$term in AP: $S_{n} = \frac{n}{2} [2a + (n-1)d ]$
Where, $a = $first term, $n= $number of terms, $d= $ common difference
$a_{1} + a_{2} + a_{3} + \ldots + a_{3n} = 1830 $
$\Rightarrow \frac{3n}{2} [ \, 2(3) + (3n-1)4 ] = 1830$
$\Rightarrow \frac{3n}{2} [ \, 6 + 12n – 4 ] = 1830$
$\Rightarrow \frac{3n}{2} [ \, 12n + 2 ] = 1830$
$\Rightarrow n(6n + 1) = 610$
$\Rightarrow 6n^{2} + n -610$
$\Rightarrow n = \frac {-1\pm \sqrt{1-4(6)(-610)}}{12}$
$\Rightarrow n = \frac {-1\pm \sqrt{1+14640}}{12}$
$\Rightarrow n = \frac {-1\pm \sqrt{14641}}{12}$
$\Rightarrow n = \frac {-1\pm 121}{12}$
$\Rightarrow n = \frac {-1-121}{12} = \frac {-122}{12}= \frac {-61}{6}\quad (\text{The number of terms can't be negative})$
$\Rightarrow n = \frac {-1+121}{12} = \frac {120}{12}= 10$
$\Rightarrow \boxed {n = 10}$
Now, $a_{n} = a + (n-1)d$
$\Rightarrow a_{10} = 3 + (10-1)4$
$\Rightarrow \boxed { a_{10} = 39 }$
Now, $m( a_{1} + a_{2} + \ldots + a_{n} ) > 1830$
$\Rightarrow m( a_{1} + a_{2} + \ldots + a_{10} ) > 1830$
$\Rightarrow m( 3 + 7 + \ldots + 39) > 1830 $
$\Rightarrow m\times \frac{10}{2} [ \, 2(3) + (10-1)4 ] > 1830$
$\Rightarrow \frac{10m}{2} (6 + 36) > 1830$
$\Rightarrow \frac{420m}{2} > 1830$
$\Rightarrow 210m> 1830$
$\Rightarrow m> \frac{183}{21}$
$\Rightarrow m> 8.71$
$\Rightarrow m = \lceil 8.71 \rceil$
$\Rightarrow \boxed { m_{\text{min}} = 9 }$
$\therefore$ The smallest positive integer $m = 9.$
Correct Answer $: \text{B}$