Solution:

Let the common difference be $d.$ Since the first term is $4$ and both $24$ and $34$ appear, $d$ divides $24 - 4 = 20$ and $34 - 4 = 30,$ so $d$ divides $\gcd(20, 30) = 10.$ The difference must be positive to reach $24$ and $34$ from $4,$ so $d \in \{1, 2, 5, 10\}$ (and each of these works, since $d \mid 20$ and $d \mid 30$ put both targets in the sequence).

The $10$th term is $4 + 9d,$ so the requested sum is $$\sum_{d \in \{1,2,5,10\}} (4 + 9d) = 4 \cdot 4 + 9(1 + 2 + 5 + 10) = 16 + 162 = 178.$$