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

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

Use the given ratio: $\frac{S_5}{S_9}=\frac{5}{17}.$

Now, $S_5=\frac{5}{2}(2a+4d)=5(a+2d),$ $S_9=\frac{9}{2}(2a+8d)=9(a+4d).$

So, $\frac{5(a+2d)}{9(a+4d)}=\frac{5}{17}.$

Cancel $5$ from both sides: $\frac{a+2d}{9(a+4d)}=\frac{1}{17}.$

Cross-multiplying, $17(a+2d)=9(a+4d).$

$17a+34d=9a+36d$ $8a=2d$ $d=4a.$

Since all terms are natural numbers, take $a=k, \quad d=4k$ for some natural number $k$.

Now find $a_{15}$: $a_{15}=a+14d=k+14(4k)=57k.$

Given $110<a_{15}<120,$ so $110<57k<120.$

Checking natural number values of $k$:

• If $k=1$, then $57k=57$ (too small)
• If $k=2$, then $57k=114$ (fits)
• If $k\ge 3$, then $57k\ge 171$ (too large)

Hence, $k=2.$

Therefore, $a=2, \quad d=8.$

Now compute the sum of first 10 terms: $S_{10}=\frac{10}{2}[2a+9d]$ $=5[2(2)+9(8)]$ $=5(4+72)$ $=5\times 76=380.$

Therefore, the required sum is $\boxed{380}.$

Option check:

• A: $290$ ❌
• B: $380$ ✅
• C: $460$ ❌
• D: $510$ ❌