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

We use:

1. $n$th term: $T_n=a+(n-1)d$

2. Sum of first $n$ terms: $S_n=\frac{n}{2}[2a+(n-1)d]$

Step 1: Use the given ratio of the 10th and 5th terms

Given: $\frac{T_{10}}{T_5}=\frac{15}{7}$

Now, $T_{10}=a+9d, \qquad T_5=a+4d$

So, $\frac{a+9d}{a+4d}=\frac{15}{7}$

Cross-multiplying: $7(a+9d)=15(a+4d)$ $7a+63d=15a+60d$ $3d=8a$ $a=\frac{3d}{8}$

Step 2: Use $S_{10}=390$

We know: $S_{10}=\frac{10}{2}[2a+9d]=390$ $5(2a+9d)=390$ $2a+9d=78$

Substitute $a=\frac{3d}{8}$: $2\left(\frac{3d}{8}\right)+9d=78$ $\frac{3d}{4}+9d=78$ $\frac{3d+36d}{4}=78$ $\frac{39d}{4}=78$ $39d=312$ $d=8$

Then, $a=\frac{3d}{8}=3$

Step 3: Compute $S_{15}-S_5$

Instead of computing separately, note that: $S_{15}-S_5$ represents the sum of terms from the 6th to the 15th term.

Now calculate directly:

$S_{15}=\frac{15}{2}[2a+14d]$ $=\frac{15}{2}[2(3)+14(8)]$ $=\frac{15}{2}[6+112]$ $=\frac{15}{2}\cdot 118$ $=15\cdot 59=885$

Also, $S_5=\frac{5}{2}[2a+4d]$ $=\frac{5}{2}[2(3)+4(8)]$ $=\frac{5}{2}[6+32]$ $=\frac{5}{2}\cdot 38=95$

Therefore, $S_{15}-S_5=885-95=790$

Step 4: Match with the options

$S_{15}-S_5=790$

So the correct option is:

C: 790

Step 5: Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

They match.