Find the sum

The formula for the sum of terms of an arithmetic progression is $s = \frac{n}{2}(a_1 + a_n)$

Considering the first five terms, we have

$30 = \frac{5}{2}(a_1 + a_5)$

After simplifying, it becomes

$a_1+a_5 = 12$

But we know that

$a_5 = a_1 + 4d$

After substituting and simplifying, we get

$2a_1+4d=12$ $\color{#D61F06}(1)$

Considering the first $10$ terms, we have

$s_{(1-10)} = 30 + 80 = 110$

$110 = \frac{10}{2}(a_1+a_{10})$

After simplifying, it becomes

$a_1+a_{10}=22$

But we know that

$a_{10}=a_1+9d$

After substituting and simplifying, we get

$2a_1+9d=22$ $\color{#D61F06}(2)$

Subtract $\color{#D61F06}(1)$ from $\color{#D61F06}(2)$ , we get

$5d = 10$

$d = 2$

Solving for $a_1$ by substituting $2$ for $d$ in $\color{#D61F06}(1)$ or $\color{#D61F06}(2)$ , we get

$2a_1+4d=12$ $\implies$ $2a_1+4(2)=12$ $\implies$ $a_1 = 2$

or

$2a_1+9d=22$ $\implies$ $2a_1+9(2)=22$ $\implies$ $a_1 = 2$

Solving for $a_{100}$ and $a_{150}$ , we have

$a_{100} = a_1 + 99d = 2 + 99(2) = 200$

$a_{150} = a_1 + 149d = 2 + 149(2) = 300$

Solving for the sum of $a_{100}$ to $a_{150}$ , we have

$s_{(100-150)} = \frac{51}{2}(200+300) =$ $\boxed{\large\color{#20A900}12750}$