#### Answer

-765

#### Work Step by Step

$\sum ^{19}_{k=5}\left( -3-4k\right) =\dfrac {\left( a_{5}+a_{19}\right) \times n}{2}\Rightarrow a_{5}=-3-4\times 5=-23;a_{19}=-3-4\times 19=-79\Rightarrow S=\dfrac {15\times \left( -23+\left( -79\right) \right) }{2}=15\times \left( -51\right) =-765$