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 $n=20$: $S_{20}=\frac{20}{2}[2a+19d]=10(2a+19d)=790$ $2a+19d=79 \quad ...(1)$

   For $n=10$: $S_{10}=\frac{10}{2}[2a+9d]=5(2a+9d)=145$ $2a+9d=29 \quad ...(2)$

4. Subtract (2) from (1): $10d=50 \Rightarrow d=5.$

5. Substitute into (2): $2a+9(5)=29$ $2a+45=29$ $2a=-16 \Rightarrow a=-8.$

6. Now find $S_{15}$ and $S_5$.

   =\frac{15}{2}[-16+70] =\frac{15}{2}\cdot 54 =15\cdot 27=405.$$ $$S_5=\frac{5}{2}[2(-8)+4(5)] =\frac{5}{2}[-16+20] =\frac{5}{2}\cdot 4=10.$$

7. Therefore, $S_{15}-S_5=405-10=395.$

8. Checking options:

• A: $405$ ❌
• B: $390$ ❌
• C: $410$ ❌
• D: $395$ ✅

Hence, the correct answer is D.