What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$? What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$? 
This is what I did: $a_9=a_1+8d=35$ 
$S_{17}=\frac{17}{2}(a_1+a_{17})=\frac{17}{2}(a_1+a_1+16d)=\frac{17}{2}(2a_1+16d)=\frac{17}{2}\cdot 70= 595$
This solution is correct, however I don't understand the solution given in the book: 

$a_1+a_2+...+a_{17}=17a_9=17\cdot 35=595$

How did they get $a_1+a_2+...+a_{17}=17a_9$?

Dang, @gt6989b is right:
\begin{align}
a_1 + a_{17} &= a_1 + (a_1 + 16 d) = 2 a_1 + 16 d = 2(a_1 + 8 d) = 2 a_9 \\
a_2 + a_{16} &= (a_1 + d) + (a_1 + 15 d)  = 2 a_1 + 16 d = 2 a_9 \\
& \vdots \\
a_8 + a_{10} &= (a_1 + 7d) + (a_1 + 9 d)  = 2 a_1 + 16 d = 2 a_9
\end{align}
so 
$$
\sum_{i=1}^{17} a_i = 8 \cdot 2 a_9 + a_9 = 17 a_9
$$
