1. Let the arithmetic progression have first term $a$ and common difference $d$.

2. Sum of first $n$ terms of an AP is $S_n=\frac{n}{2}\left[2a+(n-1)d\right].$

3. Use the given values.

   For $S_{40}=1030$: $\frac{40}{2}[2a+39d]=1030$ $20(2a+39d)=1030$ $2a+39d=\frac{1030}{20}=\frac{103}{2}. \qquad (1)$

   For $S_{12}=57$: $\frac{12}{2}[2a+11d]=57$ $6(2a+11d)=57$ $2a+11d=\frac{57}{6}=\frac{19}{2}. \qquad (2)$

4. Subtract (2) from (1): $28d=\frac{103}{2}-\frac{19}{2}=\frac{84}{2}=42$ $d=\frac{42}{28}=\frac{3}{2}.$

5. Put $d=\frac{3}{2}$ in (2): $2a+11\cdot\frac{3}{2}=\frac{19}{2}$ $2a+\frac{33}{2}=\frac{19}{2}$ $2a=-\frac{14}{2}=-7$ $a=-\frac{7}{2}.$

6. Now compute $S_{30}-S_{10}$.

   A useful identity is: $S_{30}-S_{10}=\text{sum of terms from 11th to 30th}.$

   First find $S_{30}$: $S_{30}=\frac{30}{2}[2a+29d]$ $=15\left[-7+29\cdot\frac{3}{2}\right]$ $=15\left[\frac{-14+87}{2}\right]$ $=15\cdot\frac{73}{2}=\frac{1095}{2}.$

   Next find $S_{10}$: $S_{10}=\frac{10}{2}[2a+9d]$ $=5\left[-7+9\cdot\frac{3}{2}\right]$ $=5\left[\frac{-14+27}{2}\right]$ $=5\cdot\frac{13}{2}=\frac{65}{2}.$

   Therefore, $S_{30}-S_{10}=\frac{1095}{2}-\frac{65}{2}=\frac{1030}{2}=515.$

7. Checking options:

   • A: $525$ ❌
   • B: $505$ ❌
   • C: $510$ ❌
   • D: $515$ ✅

Hence, the correct answer is Option D.