Required sum = $\left(2 + 4 + 6 + ... + 30\right)$

This is an A.P. in which $ a $ = 2, $ d $ = $\left(4 - 2\right)$ = 2 and $ l $ = 30.

Let the number of terms be $ n $. Then,

t_n = 30 $\Rightarrow$    $ a $ + $\left( n - 1\right)$$ d $ = 30

$\Rightarrow$ 2 + $\left( n - 1\right)$ x 2 = 30

$\Rightarrow n $ - 1 = 14

$\Rightarrow n $ = 15

$\therefore$S_n =$ \dfrac{n}{2} $$\left(a + l\right )$=$ \dfrac{15}{2} $ x $\left(2 + 30\right)$   = 240.