(c)

Let the number of houses be $x, x+2, x+4, x+6, x+8, x+10, \ldots$ 6 th number of house is $a$.

$\because x+10=a \Rightarrow x=a-10$

$\therefore x > 10$

Now, $\quad S_n=\frac{n}{2}(2 x+(n-1) 2)$

$S_n=n(x+n-1)$

$\Rightarrow 170=n(a-10+n-1)$

$\Rightarrow n^2+(a-11) n-170=0$

$\Rightarrow n=-(a-11) \pm \sqrt{(a-11)^2+680}$

$\Rightarrow \quad n=\frac{(11-a) \pm \sqrt{(a-11)^2+680}}{2}$

$n \geq 6$

$\Rightarrow \quad n=\frac{(11-a) \pm \sqrt{(a-11)^2+680}}{2} \geq 6$

$\Rightarrow \quad a \leq \frac{800}{24} \leq 33.33$

$\because \quad 12 \leq a \leq 32$

$a=12,14,16,18, \ldots$

When, $a=18, n=10$, then $S_n=170$ $\because \quad a=18$