The sum of the first 7 terms of an AP is 182.

Let a be the first term and d be the common difference of the AP.

$\therefore S_{7}=182$

$\Rightarrow \frac{7}{2}(2 a+6 d)=182 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$\Rightarrow a+3 d=26 \quad \ldots(1)$

Also,

$a_{4}: a_{17}=1: 5$                    (Given)

$\Rightarrow \frac{a+3 d}{a+16 d}=\frac{1}{5} \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 5 a+15 d=a+16 d$

$\Rightarrow d=4 a$                                  $\cdots \cdots(2)$

Solving (1) and (2), we get

$a+3 \times 4 a=26$

$\Rightarrow 13 a=26$

$\Rightarrow a=2$

Putting a = 2 in (2), we get

$d=4 \times 2=8$

Hence, the required AP is 2, 10, 18, 26, ... .