Let a₁, a₂,……..a_3n be an arithmetic progression with a₁ = 3 and a₂ = 7. If a₁ + a₂ + ….+a_3n = 1830, then what is the smallest positive integer m such that m (a₁ + a₂ + …. + a_n) > 1830?

The correct option is B.

Given the first term of the arithmetic progression (AP) is $a_1=3$ and the second term is $a_2=7$.
The common difference $d$ of the AP is:
$d=a_2-a_1=7-3=4$

The sum of the first $3n$ terms is given as 1830:

Substitute $a_1=3$ and $d=4$ into the equation:
$\frac{3n}{2}\left[2\right(3)+(3n-1\left)4\right]=1830$
$\frac{3n}{2}[6+12n-4]=1830$
$\frac{3n}{2}[12n+2]=1830$
$3n(6n+1)=1830$
$n(6n+1)=610$
$6n^2+n-610=0$

Solving the quadratic equation:
$(6n+61)(n-10)=0$
Since $n$ must be a positive integer, we have $n=10$.

Now, let's find the sum of the first $n$ terms, i.e., $S_{10}$:
$S_n=a_1+a_2+\dots +a_n$
$S_{10}=\frac{10}{2}\left[2\right(3)+(10-1\left)4\right]$
$S_{10}=5[6+36]=5\left(42\right)=210$

We need to find the smallest positive integer $m$ such that:
$m\left(S_n\right)>1830$
$m\left(210\right)>1830$
$m>\frac{1830}{210}$
$m>8.71$
Thus, the smallest positive integer $m$ satisfying this inequality is 9.