We can insert the seven numbers in such a way that, so we get the arithmetic progression including $2$ and $34.$
We get the arithmetic progression $2,6,10,14,18,22,26,30,34$
The inserted seven numbers are $6,10,14,18,22,26,30$
Here, $a = 6,d = 10-6 = 4,l = 30,n = 7$
$\therefore$ The sum of the arithmetic progression $S_{7} = \frac{7}{2}\left(6 + 30 \right) = 7 \cdot 18 = 126.$
So, the correct answer is $(C).$