Let a & d are first term and common diff of an AP.

$$\begin{aligned} & \mathrm{S}_{40}=\frac{40}{2}[2 \mathrm{a}+39 \mathrm{~d}]=1030 \quad\text{..... (1)}\\ & \mathrm{~S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57 \quad\text{..... (2)} \end{aligned}$$

by (1) & (2)

$$\begin{aligned} & \mathrm{a}=-\frac{7}{2} \quad \mathrm{~d}=\frac{3}{2} \\ & \therefore \mathrm{~S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}] \\ & =20 \mathrm{a}-390 \mathrm{~d} \\ & =515 \end{aligned}$$