jmc

Taking $n=1,$ we get $pq+r=14.$ Also, from the formula $a_n=24-5a_{n-1},$ $$p\cdot q^n+r=24-5(p\cdot q^{n-1}+r)=24-5p\cdot q^{n-1}-5r.$$We can write this as $$pq\cdot q^{n-1}+r=24-5p\cdot q^{n-1}-5r.$$Then we must have $pq=-5p$ and $r=24-5r.$ Hence, $6r=24,$ so $r=4.$

From $pq+5p=0,$ $p(q+5)=0,$ so $p=0$ or $q=-5.$ If $p=0,$ then $r=14,$ contradiction, so $q=-5.$ Then $$-5p+4=14,$$whence $p=-2.$ Therefore, $p+q+r=(-2)+(-5)+4=\boxed{-3}.$