The first and the last terms of an AP are 17 and 350 respectively.

Suppose there are n terms in the AP.

Here, a = 17, d = 9 and l = 350

$\therefore a_{n}=350$

$\Rightarrow 17+(n-1) \times 9=350 \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 9 n+8=350$

$\Rightarrow 9 n=350-8=342$

$\Rightarrow n=38$

Thus, there are 38 terms in the AP.

$\therefore S_{38}=\frac{38}{2}(17+350) \quad\left[S_{n}=\frac{n}{2}(a+l)\right]$

$=19 \times 367$

$=6973$

Hence, the required sum is 6973.