Answer: 178

Small Hint:

If the common difference is $d,$ then $24$ and $34$ both being terms forces $d$ to divide both $20$ and $30.$

Big Hint:

The common difference must be a positive divisor of $10;$ add up the $10$th terms $4 + 9d$ over all such $d.$

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 $20$ and $30$ are both divisible by $d,$ which puts both targets in the sequence).

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