The sum of first $26$terms of an A.P  $a_1,a_2,a_3,\dots ,$   If $a_2+a_6+a_9+a_{18}+a_{21}+a_{25}=165,$ is 

We  have

$a_1+a_{26}=a_2+a_{25}=a_3+a_{24}=\dots =a_9+a_{18}$

Thus, 

$\begin{array}{l}{a}_2+a_6+a_9+a_{18}+a_{21}+a_{25}=165\\ \Rightarrow \left(a_2+a_{25}\right)+\left(a_6+a_{21}\right)+\left(a_9+a_{18}\right)=165\\ \Rightarrow 3\left(a_1+a_{26}\right)=165\Rightarrow a_1+a_{26}=55\\ \therefore a_1+a_2+\dots +a_{26}=\frac{26}{2}\left(a_1+a_{26}\right)\\ =13\times 55=715\end{array}$